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TRANSCRIPT
Chapter
Expanding binomial productsPerfect squares and difference of perfect squaresFactorising algebraic expressionsFactorising the difference of two squaresFactorisation by groupingFactorising quadratic trinomials (Extending)Factorising trinomials of the form ax2 + bx + c (Extending)Simplifying algebraic fractions: multiplication and divisionSimplifying algebraic fractions: addition and subtractionFurther simpli� cation of algebraic fractions (Extending)Equations with algebraic fractions (Extending)
8A8B
8C8D8E8F
8G
8H
8I
8J
8K
N U M B E R A N D A L G E B R A
Patterns and algebraApply the distributive law to the expansion of algebraic expressions, including binomials, and collect like terms where appropriate
16x16 32x32
What you will learn
Australian curriculum
8 Algebraic techniques
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The distance (x units) of an object from the top of a building after it has been dropped (where air resistance is negligible) can be found using the formula x = ut + 1 2 at 2 where u is the initial velocity of the object, t the time since the object has been dropped and a the acceleration due to gravity, which is approximately equal to –9.8 m/s2. When an object is dropped it has an initial velocity of 0 m/s, so the distance the object has fallen becomes x = –4.9t 2. Using algebra, the distance from the building after t seconds can be found or the time taken to reach ground level could be calculated. If
_
the object is instead dropped from a hot air balloon ascending at 10 m/s, the object � rst travels in an upward direction. Its distance (x metres) above or below the height of the balloon from when the object is dropped can be found using x = 10t – 4.9t 2. Knowing the time taken for the object to reach the ground, we could again use algebra to � nd factors, such as the height of the balloon, the greatest height reached by the object and the time taken for the object to return to the height from which it was released.
• Chapter pre-test• Videos of all worked
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Online resources
Free falling
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488 488 Chapter 8 Algebraic techniques
8A Expanding binomial productsA binomial is an expression with
two terms such as x + 5 or x2 + 3.
You will recall from Chapter 2 that
we looked at the product of a single
term with a binomial expression, e.g.
2(x – 3) or x(3x – 1). The product of
two binomial expressions can also be
expanded using the distributive law.
This involves multiplying every term
in one expression by every term in the
other expression.
Expanding the product of two binomial expressions can be applied toproblems involving the expansion of rectangular areas such as a farmer’spaddock.
Let’s start: Rectangular expansions
If (x + 1) and (x + 2) are the side lengths of a rectangle as shown, the total area can be found as an
expression in two different ways.
• Write an expression for the total area of the rectangle using length = (x + 2) and
width = (x + 1).
• Now find the area of each of the four parts of the rectangle and combine to give
an expression for the total area.
• Compare your two expressions above and complete this equation:
(x + 2)( ) = x2 + + .
• Can you explain a method for expanding the left-hand side to give the right-hand
side?
x
x
1
2
Keyideas
Expanding binomial products uses the distributive law.
(a + b)(c + d) = a(c + d) + b(c + d)
= ac + ad + bc + bd
Diagrammatically (a + b)(c + d) = ac + ad + bc + bd
a
ac bc
ad bdd
c
b
For example: (x + 1)(x + 5) = x2 + 5x + x + 5
= x2 + 6x + 5
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Number and Algebra 489 489
Example 1 Expanding binomial products
Expand the following.
(x + 3)(x + 5)a (x – 4)(x + 7)b
(2x – 1)(x – 6)c (5x – 2)(3x + 7)d
SOLUTION EXPLANATION
a (x + 3)(x + 5) = x2 + 5x + 3x + 15
= x2 + 8x + 15
Use the distributive law to expand the brackets
and then collect the like terms 5x and 3x.
b (x − 4)(x + 7) = x2 + 7x – 4x – 28
= x2 + 3x – 28
After expanding to get the four terms, collect
the like terms 7x and – 4x.
c (2x − 1)(x − 6) = 2x2 – 12x – x + 6
= 2x2 – 13x + 6
Remember 2x× x = 2x2 and –1 × (–6) = 6.
d (5x − 2)(3x + 7) = 15x2 + 35x – 6x – 14
= 15x2 + 29x – 14
Recall 5x× 3x = 5 × 3 × x× x = 15x2.
Exercise 8A
1 The given diagram shows the area (x + 2)(x + 3).
a Write down an expression for the area of each of the four regions inside the
rectangle.
b Copy and complete:
(x + 2)(x + 3) = + 3x + + 6
= + 5x +
x
x
2
3
2 The given diagram shows the area (2x + 3)(x + 1).
a Write down an expression for the area of each of the four regions
inside the rectangle.
b Copy and complete:
(2x + 3)( ) = 2x2 + + 3x +
= + +
2x
x
1
3
UNDE
RSTA
NDING
—1–3 3(½)
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490 490 Chapter 8 Algebraic techniques
8A3 Copy and complete these expansions.
(x + 1)(x + 5) = + 5x + + 5
= + 6x +
a (x – 3)(x + 2) = + – 3x –
= – x –
b
(3x – 2)(7x + 2) = + 6x – –
= – +
c (4x – 1)(3x – 4) = – – 3x +
= – 19x +
d UNDE
RSTA
NDING
4Example 1a Expand the following.
(x + 2)(x + 5)a (b + 3)(b + 4)b (t + 8)(t + 7)c(p + 6)(p + 6)d (x + 9)(x + 6)e (d + 15)(d + 4)f(a + 1)(a + 7)g (y + 10)(y + 2)h (m + 4)(m + 12)i
5Example 1b,c,d Expand the following.
(x + 3)(x – 4)a (x + 5)(x – 2)b (x + 4)(x – 8)c(x – 6)(x + 2)d (x – 1)(x + 10)e (x – 7)(x + 9)f(x – 2)(x + 7)g (x – 1)(x – 2)h (x – 4)(x – 5)i(4x + 3)(2x + 5)j (3x + 2)(2x + 1)k (3x + 1)(5x + 4)l(2x – 3)(3x + 5)m (8x – 3)(3x + 4)n (3x – 2)(2x + 1)o(5x + 2)(2x – 7)p (2x + 3)(3x – 2)q (4x + 1)(4x – 5)r(3x – 2)(6x – 5)s (5x – 2)(3x – 1)t (7x – 3)(3x – 4)u
6 Expand these binomial products.
(a + b)(a + c)a (a – b)(a + c)b (b – a)(a + c)c(x – y)(y – z)d (y – x)(z – y)e (1 – x)(1 + y)f(2x + y)(x – 2y)g (2a + b)(a – b)h (3x – y)(2x + y)i(2a – b)(3a + 2)j (4x – 3y)(3x – 4y)k (xy – yz)(z + 3x)l
FLUE
NCY
4–6(½)4–5(½) 4–6(½)
7 A room in a house with dimensions
4 m by 5 m is to be extended. Both
the length and width are to be
increased by x m.
a Find an expanded expression for
the area of the new room.
b If x = 3:
find the area of the new roomiby how much has the area
increased?
ii
PROB
LEM-SOLVING
8, 97 7, 8
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Number and Algebra 491 491
8A8 A picture frame 5 cm wide has a length which is twice the
width x cm.
a Find an expression for the total area of the frame and
picture.
b Find an expression in expanded form for the area of the
picture only.
2x cm5 cm
x cmPicture
9 The outside edge of a path around a rectangular swimming pool
is 15 m long and 10 m wide. The path is x metres wide.
a Find an expression for the area of the pool in expanded
form.
b Find the area of the pool if x = 2. 15 m
10 m
x m
Pool
PROB
LEM-SOLVING
10 Write the missing terms in these expansions.
(x + 2)(x + ) = x2 + 5x + 6a (x + )(x + 5) = x2 + 7x + 10b(x + 1)(x + ) = x2 + 7x +c (x + )(x + 9) = x2 + 11x +d(x + 3)(x – ) = x2 + x –e (x – 5)(x + ) = x2 – 2x –f(x + 1)( + 3) = 2x2 + +g ( – 4)(3x – 1) = 9x2 – +h(x + 2)( + ) = 7x2 + + 6i ( – )(2x – 1) = 6x2 – + 4j
11 Consider the binomial product (x + a)(x + b). Find the possible integer values of a and b if:
(x + a)(x + b) = x2 + 5x + 6a (x + a)(x + b) = x2 – 5x + 6b(x + a)(x + b) = x2 + x – 6c (x + a)(x + b) = x2 – x – 6d
REAS
ONING
10(½), 1110(½) 10(½), 11
Trinomial expansions
12 Using the distributive law (a + b)(c + d + e) = ac + ad + ae + bc + bd + be.
Use this knowledge to expand and simplify these products. Note: x× x2 = x3.
(x + 1)(x2 + x + 1)a (x – 2)(x2 – x + 3)b(2x – 1)(2x2 – x + 4)c (x2 – x + 1)(x + 3)d(5x2 – x + 2)(2x – 3)e (2x2 – x + 7)(4x – 7)f(x + a)(x2 – ax + a)g (x – a)(x2 – ax – a2)h(x + a)(x2 – ax + a2)i (x – a)(x2 + ax + a2)j
13 Now try to expand (x + 1)(x + 2)(x + 3).
ENRICH
MEN
T12, 13— —
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492 492 Chapter 8 Algebraic techniques
8B Perfect squares and difference of perfect squares
We know that 22 = 4, 152 = 225, x2 and (a + b)2 are all examples of perfect squares. To expand (a + b)2
we multiply (a + b) by (a + b) and use the distributive law:
(a + b)2 = (a + b)(a + b)
= a2 + ab + ba + b2
= a2 + 2ab + b2
A similar result is obtained for the
square of (a – b):
(a – b)2 = (a – b)(a – b)
= a2 – ab – ba + b2
= a2 – 2ab + b2
Another type of expansion involves
the case that deals with the product
of the sum and difference of the
same two terms. The result is the
difference of two perfect squares:
Binomial products can be used to calculate the most efficient way to cut theshapes required for a fabrication out of a metal sheet.
(a + b)(a – b) = a2 – ab + ba – b2
= a2 – b2 (since ab = ba, the two middle terms cancel each other out.)
Let’s start: Seeing the pattern
Using (a + b)(c + d) = ac + ad + bc + bd, expand and simplify the binomial products in the two sets
below.
Set A(x + 1)(x + 1) = x2 + x + x + 1
=
(x + 3)(x + 3) =
=
(x – 5)(x – 5) =
=
Set B(x + 1)(x – 1) = x2 – x + x – 1
=
(x – 3)(x + 3) =
=
(x – 5)(x + 5) =
=
• Describe what patterns you see in both sets of expansions above.
• Generalise your observations by completing the following expansions.
(a + b)(a + b) = a2 + + +
= a2 + +
(a – b)(a – b) =
=
A (a + b)(a – b) = a2 – + –
=
B
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Number and Algebra 493 493
Keyideas
32 = 9, a2, (2y)2, (x – 1)2 and (3 – 2y)2 are all examples of perfect squares.Expanding perfect squares
(a + b)2 = (a + b)(a + b)
= a2 + ab + ba + b2
= a2 + 2ab + b2
•
(a – b)2 = (a – b)(a – b)
= a2 – ab – ba + b2
= a2 – 2ab + b2
•
Difference of perfect squares (DOPS)
• (a + b)(a – b) = a2 – ab + ba – b2
= a2 – b2
• (a – b)(a + b) also expands to a2 – b2
• The result is a difference of two perfect squares.
Example 2 Expanding perfect squares
Expand each of the following.
(x – 2)2a (2x + 3)2b
SOLUTION EXPLANATION
a (x – 2)2 = (x – 2)(x – 2)
= x2 – 2x – 2x + 4
= x2 – 4x + 4
Alternative solution:
(x – 2)2 = x2 – 2 × x× 2 + 22
= x2 – 4x + 4
Write in expanded form.
Use the distributive law.
Collect like terms.
Expand using (a – b)2 = a2 – 2ab + b2 where
a = x and b = 2.
b (2x + 3)2 = (2x + 3)(2x + 3)
= 4x2 + 6x + 6x + 9
= 4x2 + 12x + 9
Alternative solution:
(2x + 3)2 = (2x)2 + 2 × 2x× 3 + 32
= 4x2 + 12x + 9
Write in expanded form.
Use the distributive law.
Collect like terms.
Expand using (a + b)2 = a2 + 2ab + b2 where
a = 2x and b = 3. Recall (2x)2 = 2x× 2x = 4x2.
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494 494 Chapter 8 Algebraic techniques
Example 3 Forming a difference of perfect squares
Expand and simplify the following.
(x + 2)(x – 2)a (3x – 2y)(3x + 2y)b
SOLUTION EXPLANATION
a (x + 2)(x – 2) = x2 – ��2x + ��2x – 4
= x2 – 4
Expand using the distributive law.
–2x + 2x = 0.
Alternative solution:
(x + 2)(x – 2) = (x)2 – (2)2
= x2 – 4
(a + b)(a – b) = a2 – b2. Here a = x and b = 2.
b (3x – 2y)(3x + 2y) = 9x2 + ��6xy – ��6xy – 4y2
= 9x2 – 4y2Expand using the distributive law.
6xy – 6xy = 0.
Alternative solution:
(3x – 2y)(3x + 2y) = (3x)2 – (2y)2
= 9x2 – 4y2(a + b)(a – b) = a2 – b2 with a = 3x and b = 2y
here.
Exercise 8B
1 Complete these expansions.
(x + 3)(x + 3) = x2 + 3x + +
=
a (x + 5)(x + 5) = x2 + 5x + +
=
b
(x – 2)(x – 2) = x2 – 2x – +
=
c (x – 7)(x – 7) = x2 – 7x – +
=
d
2 a Substitute the given value of b into x2 + 2bx + b2 and simplify.
b = 3i b = 11ii b = 15iii
b Substitute the given value of b into x2 – 2bx + b2 and simplify.
b = 2i b = 9ii b = 30iii
3 Complete these expansions.
(x + 4)(x – 4) = x2 – 4x + –
=
a (x – 10)(x + 10) = x2 + 10x – –
=
b
(2x – 1)(2x + 1) = 4x2 + – –
=
c (3x + 4)(3x – 4) = 9x2 – + –
=
d
UNDE
RSTA
NDING
—1–3 1(½), 3(½)
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Number and Algebra 495 495
8B4Example 2a Expand each of the following perfect squares.
(x + 1)2a (x + 3)2b (x + 2)2c (x + 5)2d(x + 4)2e (x + 9)2f (x + 7)2g (x + 10)2h(x – 2)2i (x – 6)2j (x – 1)2k (x – 3)2l(x – 9)2m (x – 7)2n (x – 4)2o (x – 12)2p
5Example 2b Expand each of the following perfect squares.
(2x + 1)2a (2x + 5)2b (3x + 2)2c(3x + 1)2d (5x + 2)2e (4x + 3)2f(7 + 2x)2g (5 + 3x)2h (2x – 3)2i(3x – 1)2j (4x – 5)2k (2x – 9)2l(3x + 5y)2m (2x + 4y)2n (7x + 3y)2o(6x + 5y)2p (4x – 9y)2q (2x – 7y)2r(3x – 10y)2s (4x – 6y)2t (9x – 2y)2u
6 Expand each of the following perfect squares.
(3 – x)2a (5 – x)2b (1 – x)2c(6 – x)2d (11 – x)2e (4 – x)2f(7 – x)2g (12 – x)2h (8 – 2x)2i(2 – 3x)2j (9 – 2x)2k (10 – 4x)2l
7Example 3a Expand and simplify the following to form a difference of perfect squares.
(x + 1)(x – 1)a (x + 3)(x – 3)b (x + 8)(x – 8)c(x + 4)(x – 4)d (x + 12)(x – 12)e (x + 11)(x – 11)f(x – 9)(x + 9)g (x – 5)(x + 5)h (x – 6)(x + 6)i(5 – x)(5 + x)j (2 – x)(2 + x)k (7 – x)(7 + x)l
8Example 3b Expand and simplify the following.
(3x – 2)(3x + 2)a (5x – 4)(5x + 4)b (4x – 3)(4x + 3)c(7x – 3y)(7x + 3y)d (9x – 5y)(9x + 5y)e (11x – y)(11x + y)f(8x + 2y)(8x – 2y)g (10x – 9y)(10x + 9y)h (7x – 5y)(7x + 5y)i(6x – 11y)(6x + 11y)j (8x – 3y)(8x + 3y)k (9x – 4y)(9x + 4y)l
FLUE
NCY
4–8(½)4–7(½) 4–8(½)
9 Lara is x years old and her two best friends are (x – 2) and (x + 2) years old.
a Write an expression for:
the square of Lara’s ageithe product of the ages of Lara’s best friends (in expanded form).ii
b Are the answers from parts a i and ii equal? If not, by how much do they differ?
PROB
LEM-SOLVING
9, 109 9, 10
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496 496 Chapter 8 Algebraic techniques
8B10 A square piece of tin of side length 20 cm has four
squares of side length x cm removed from each corner.
The sides are folded up to form a tray. The centre
square forms the tray base.
a Write an expression for the side length of the base
of the tray.
b Write an expression for the base of the tray. Expand
your answer.
c Find the area of the tray base if x = 3.
d Find the volume of the tray if x = 3.
Traybase
x cm
x cm
20 cm
20 cm PROB
LEM-SOLVING
11 Four tennis courts are arranged as shown with a square storage area in the centre. Each court area
has the same dimensions a× b.
a Write an expression for the side length of the total area.
b Write an expression for the total area.
c Write an expression for the side length of the inside storage area.
d Write an expression for the area of the inside storage area.
e Subtract your answer to part d from your answer to part b to find the
area of the four courts.
a
b
f Find the area of one court. Does your answer confirm that your answer to part e is correct?
12 A square of side length x units has one side reduced by 1 unit and the other increased by 1 unit.
a Find an expanded expression for the area of the resulting rectangle.
b Is the area of the original square the same as the area of the resulting rectangle?
Explain why/why not?
REAS
ONING
12, 1311 12, 13
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Number and Algebra 497 497
8B13 A square of side length b is removed from a square of side length a.
A B
C b
a
a Using subtraction write down an expression for the remaining area.
b Write expressions for the area of the regions:
Ai Bii Ciii
c Add all the expressions from part b to see if you get your answer from part a.
REAS
ONING
Extended expansions
14 Expand and simplify these expressions.
(x + 2)2 – 4a (2x – 1)2 – 4x2b(x + 3)(x – 3) + 6xc 1 – (x + 1)2dx2 – (x + 1)(x – 1)e (x + 1)2 – (x – 1)2f(3x – 2)(3x + 2) – (3x + 2)2g (5x – 1)2 – (5x + 1)(5x – 1)h(x + y)2 – (x – y)2 + (x + y)(x – y)i (2x – 3)2 + (2x + 3)2j(2 – x)2 – (2 + x)2k (3 – x)2 + (x – 3)2l2(3x – 4)2 – (3x – 4)(3x + 4)m 2(x + y)2 – (x – y)2n
The logical skills of algebra have applications in computer programming.
ENRICH
MEN
T
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498 498 Chapter 8 Algebraic techniques
8C Factorising algebraic expressionsThe process of factorisation is a key
step in the simplification of many
algebraic expressions and in the solution
of equations. It is the reverse process
of expansion and involves writing an
expression as a product of its factors.
expanding−−−−−−→
2(x – 3) = 2x – 6←−−−−−−factorising
Factorising is a key mathematical skill required in many diverseoccupations, such as in business, science, technology and engineering.
Let’s start: Which factorised form?
The product x(4x + 8) when expanded gives 4x2 + 8x.
• Write down three other products that when expanded give 4x2 + 8x. (Do not use fractions.)
• Which of your products uses the highest common factor of 4x2 and 8x? What is this highest common
factor?
Keyideas
When factorising expressions with common factors, take out the highest common factor (HCF).
The HCF could be:
• a number
For example: 2x + 10 = 2(x + 5)
• a pronumeral (or variable)
For example: x2 + 5x = x(x + 5)
• the product of numbers and pronumerals
For example: 2x2 + 10x = 2x(x + 5)
A factorised expression can be checked by using expansion.
For example: 2x(x + 5) = 2x2 + 10x.
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Number and Algebra 499 499
Example 4 Finding the HCF
Determine the HCF of the following.
6a and 8aba 3x2 and 6xyb
SOLUTION EXPLANATION
a 2a HCF of 6 and 8 is 2.
HCF of a and ab is a.
b 3x HCF of 3 and 6 is 3.
HCF of x2 and xy is x.
Example 5 Factorising expressions
Factorise the following.
40 – 16ba –8x2 – 12xb
SOLUTION EXPLANATION
a 40 – 16b = 8(5 – 2b) The HCF of 40 and 16b is 8. Place 8 in front
of the brackets and divide each term by 8.
b –8x2 – 12x = – 4x(2x + 3) The HCF of the terms is – 4x, including the
common negative. Place the factor in front of
the brackets and divide each term by – 4x.
Example 6 Taking out a binomial factor
Factorise the following.
3(x + y) + x(x + y)a (7 – 2x) – x(7 – 2x)b
SOLUTION EXPLANATION
a 3(x + y) + x(x + y)
= (x + y)(3 + x)
HCF = (x + y).
The second pair of brackets contains what
remains when 3(x + y) and x(x + y) are divided
by (x + y).
b (7 – 2x) – x(7 – 2x)
= 1(7 – 2x) – x(7 – 2x)
= (7 – 2x)(1 – x)
Insert 1 in front of the first bracket.
HCF = (7 – 2x).
The second bracket must contain 1 – x after
dividing (7 – 2x) and x(7 – 2x) by (7 – 2x).
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500 500 Chapter 8 Algebraic techniques
Exercise 8C
1 Write down the highest common factor (HCF) of these pairs of numbers.
8, 12a 10, 20b 5, 60c 24, 30d3, 5e 100, 75f 16, 24g 36, 72h
2 Write down the missing factor.
5 × = 5xa 7 × = 7xb 2a× = 2a2c5a× = 10a2d × 3y = –6y2e × 12x = –36x2f–3 × = 6xg –2x× = 20x2h × 7xy = –14x2yi
3 a Write down the missing factor in each part.
(x2 + 2x) = 6x2 + 12xi (2x + 4) = 6x2 + 12xii(x + 2) = 6x2 + 12xiii
b Which equation above uses the HCF of 6x2 and 12x?
c By looking at the terms left in the brackets, how do you know you have taken out the HCF?
UNDE
RSTA
NDING
—1–2(½), 3 3
4Example 4 Determine the HCF of the following.
6x and 14xya 12a and 18ab 10m and 4c 12y and 8d15t and 6se 15 and pf 9x and 24xyg 6n and 21mnh10y and 2yi 8x2 and 14xj 4x2y and 18xyk 5ab2 and 15a2bl
5Example 5a Factorise the following.
7x + 7a 3x + 3b 4x – 4c 5x – 5d4 + 8ye 10 + 5af 3 – 9bg 6 – 2xh12a + 3bi 6m + 6nj 10x – 8yk 4a – 20blx2 + 2xm a2 – 4an y2 – 7yo x – x2p3p2 + 3pq 8x – 8x2r 4b2 + 12bs 6y – 10y2t12a – 15a2u 9m + 18m2v 16xy – 48x2w 7ab – 28ab2x
6Example 5b Factorise the following by factoring out the negative sign as part of the HCF.
–8x – 4a – 4x – 2b –10x – 5yc –7a – 14bd–9x – 12e –6y – 8f –10x – 15yg – 4m – 20nh–3x2 – 18xi –8x2 – 12xj –16y2 – 6yk –5a2 – 10al–6x – 20x2m –6p – 15p2n –16b – 8b2o –9x – 27x2p
7Example 6 Factorise the following which involve a binomial common factor.
4(x + 3) + x(x + 3)a 3(x + 1) + x(x + 1)b 7(m – 3) + m(m – 3)cx(x – 7) + 2(x – 7)d 8(a + 4) – a(a + 4)e 5(x + 1) – x(x + 1)fy(y + 3) – 2(y + 3)g a(x + 2) – x(x + 2)h t(2t + 5) + 3(2t + 5)im(5m – 2) + 4(5m – 2)j y(4y – 1) – (4y – 1)k (7 – 3x) + x(7 – 3x)l
FLUE
NCY
4–8(½)4–7(½) 4–8(½)
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8C8 Factorise these mixed expressions.
6a + 30a 5x – 15b 8b + 18cx2 – 4xd y2 + 9ye a2 – 3afx2y – 4xy + xy2g 6ab – 10a2b + 8ab2h m(m + 5) + 2(m + 5)ix(x + 3) – 2(x + 3)j b(b – 2) + (b – 2)k x(2x + 1) – (2x + 1)ly(3 – 2y) – 5(3 – 2y)m (x + 4)2 + 5(x + 4)n (y + 1)2 – 4(y + 1)o
FLUE
NCY
9 Write down the perimeter of these shapes in factorised form.
2x
4
ax
6
b
5x
10
c
x
5
2
d
x + 1
1
e3
4x
f
10 The expression for the area of a rectangle is (4x2 + 8x) square units. Find an expression for its
width if the length is (x + 2) units.
11 The height, in metres, of a ball thrown in the air is given by
5t – t2, where t is the time in seconds.
a Write an expression for the ball’s height in factorised
form.
b Find the ball’s height at these times:
t = 0it = 2iit = 4iii
c How long does it take for the ball’s height to return to
0 metres? Use trial and error if required.
PROB
LEM-SOLVING
9(½), 10, 119 9, 10
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502 502 Chapter 8 Algebraic techniques
8C12 7 × 9 + 7 × 3 can be evaluated by firstly factorising to 7(9 + 3). This gives 7 × 12 = 84. Use a
similar technique to evaluate the following.
9 × 2 + 9 × 5a 6 × 3 + 6 × 9b –2 × 4 – 2 × 6c–5 × 8 – 5 × 6d 23 × 5 – 23 × 2e 63 × 11 – 63 × 8f
13 Common factors can also be removed from expressions with more than two terms.
For example: 2x2 + 6x + 10xy = 2x(x + 3 + 5y)
Factorise these expressions by taking out the HCF.
3a2 + 9a + 12a 5z2 – 10z + zyb x2 – 2xy + x2yc4by – 2b + 6b2d –12xy – 8yz – 20xyze 3ab + 4ab2 + 6a2bf
14 Sometimes we can choose to factor out a negative or a positive HCF. Both factorisations are
correct. For example:
–13x + 26 = –13(x – 2) (HCF is –13)
OR –13x + 26 = 13(–x + 2) (HCF is 13)
= 13(2 – x)
Factorise in two different ways: the first by factoring out a negative and the second by a
positive HCF.
– 4x + 12a –3x + 9b –8n + 8c –3b + 3d–5m + 5m2e –7x + 7x2f –5x + 5x2g – 4y + 22y2h–8n + 12n2i –8y + 20j –15mn + 10k –15x + 45l
REAS
ONING
13, 1412 12, 13
Factoring out a negative
15 Using the fact that a – b = –(b – a) you can factorise x(x – 2) – 5(2 – x) by following these steps.
x(x – 2) – 5(2 – x) = x(x – 2) + 5(x – 2)
= (x – 2)(x + 5)
Use this idea to factorise these expressions.
x(x – 4) + 3(4 – x)a x(x – 5) – 2(5 – x)b x(x – 3) – 3(3 – x)c3x(x – 4) + 5(4 – x)d 3(2x – 5) + x(5 – 2x)e 2x(x – 2) + (2 – x)f– 4(3 – x) – x(x – 3)g x(x – 5) + (10 – 2x)h x(x – 3) + (6 – 2x)i
ENRICH
MEN
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Number and Algebra 503 503
8D Factorising the difference of two squaresRecall that a difference of two perfect squares is formed when expanding the product of the sum and
difference of two terms. For example, (x + 2)(x – 2) = x2 – 4. Reversing this process means that a
difference of two perfect squares can be factorised into two binomial expressions of the form (a + b)
and (a – b).
Let’s start: Expanding to understand factorising
Complete the steps in these expansions then write the conclusion.
(x + 3)(x – 3) = x2 – 3x + –
= x2 –
∴ x2 – 9 = ( + )( – )
• (2x – 5)(2x + 5) = 4x2 + 10x – –
= –
∴ 4x2 – = ( + )( – )
•
(a + b)(a – b) = a2 – ab + –
= –
∴ a2 – = ( + )( – )
•
Keyideas
Factorising the difference of perfect squares (DOPS) uses the rule a2 – b2 = (a + b)(a – b).
• x2 – 16 = x2 – 42
= (x + 4)(x – 4)
• 9x2 – 100 = (3x)2 – 102
= (3x + 10)(3x – 10)
• 25 – 4y2 = 52 – (2y)2
= (5 + 2y)(5 – 2y)
First take out common factors where possible.
• 2x2 – 18 = 2(x2 – 9)
= 2(x + 3)(x – 3)
Example 7 Factorising DOPS
Factorise each of the following.
x2 – 4a 9a2 – 25b
81x2 – y2c 2b2 – 32d
(x + 1)2 – 4e
SOLUTION EXPLANATION
a x2 – 4 = x2 – 22
= (x + 2)(x – 2)
Write as a DOPS (4 is the same as 22).
Write in factorised form a2 –b2 = (a+b)(a–b).
Here a = x and b = 2.
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504 504 Chapter 8 Algebraic techniques
b 9a2 – 25 = (3a)2 – 52
= (3a + 5)(3a – 5)
Write as a DOPS. 9a2 is the same as (3a)2.
Write in factorised form.
c 81x2 – y2 = (9x)2 – y2
= (9x + y)(9x – y)
81x2 = (9x)2
Use a2 – b2 = (a + b)(a – b)
d 2b2 – 32 = 2(b2 – 16)
= 2(b2 – 42)
= 2(b + 4)(b – 4)
First, factor out the common factor of 2.
Write as a DOPS and then factorise.
e (x + 1)2 – 4 = (x + 1)2 – 22
= (x + 1 + 2)(x + 1 – 2)
= (x + 3)(x – 1)
Write as a DOPS. In a2 – b2 here, a is the
expression x + 1 and b = 2.
Write in factorised form and simplify.
Exercise 8D
1 Expand these binomial products to form a difference of perfect squares.
(x + 2)(x – 2)a (x – 7)(x + 7)b (2x – 1)(2x + 1)c(x + y)(x – y)d (3x – y)(3x + y)e (a + b)(a – b)f
2 Write the missing term. Assume it is a positive number.
( )2 = 9a ( )2 = 121b ( )2 = 81c ( )2 = 400d( )2 = 4x2e ( )2 = 9a2f ( )2 = 25b2g ( )2 = 49y2h
3 Complete these factorisations.
x2 – 16 = x2 – 42
= (x + 4)( – )
a x2 – 144 = x2 – ( )2
= ( + 12)(x – )
b
16x2 – 1 = ( )2 – ( )2
= (4x + )( – 1)
c 9a2 – 4b2 = ( )2 – ( )2
= (3a + )( – 2b)
dUN
DERS
TAND
ING
—1–3(½) 3(½)
4Example 7a Factorise each of the following.
x2 – 9a y2 – 25b y2 – 1c x2 – 64dx2 – 16e b2 – 49f a2 – 81g x2 – y2ha2 – b2i 16 – a2j 25 – x2k 1 – b2l36 – y2m 121 – b2n x2 – 400o 900 – y2p
FLUE
NCY
4–7(½)4–6(½) 4–7(½)
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8D5Example 7b,c Factorise each of the following.
4x2 – 25a 9x2 – 49b 25b2 – 4c 4m2 – 121d100y2 – 9e 81a2 – 4f 1 – 4x2g 25 – 64b2h16 – 9y2i 36x2 – y2j 4x2 – 25y2k 64a2 – 49b2l4p2 – 25q2m 81m2 – 4n2n 25a2 – 49b2o 100a2 – 9b2p
6Example 7d Factorise each of the following by first taking out the common factor.
3x2 – 108a 10a2 – 10b 6x2 – 24c4y2 – 64d 98 – 2x2e 32 – 8m2f5x2y2 – 5g 3 – 3x2y2h 63 – 7a2b2i
7Example 7e Factorise each of the following.
(x + 5)2 – 9a (x + 3)2 – 4b (x + 10)2 – 16c(x – 3)2 – 25d (x – 7)2 – 1e (x – 3)2 – 36f49 – (x + 3)2g 4 – (x + 2)2h 81 – (x + 8)2i
FLUE
NCY
8 The height above ground (in metres) of an object
thrown off the top of a building is given by 36 – 4t2
where t is in seconds.
a Factorise the expression for the height of the
object by firstly taking out the common factor.
b Find the height of the object:
initially (t = 0)iat 2 seconds (t = 2).ii
c How long does it take for
the object to hit the ground?
Use trial and error if you wish.
9 This ‘multisize’ square picture frame has side length 30 cm and can
hold a square picture with any side length less than 26 cm.
a If the side length of the picture is x cm, write an expression for:
the area of the pictureithe area of the frame (in factorised form).ii
b Use your result from part a ii to find the area of the frame if:
x = 20ithe area of the picture is 225 cm2.ii
30 cmx cm
PROB
LEM-SOLVING
8, 98 8, 9
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506 506 Chapter 8 Algebraic techniques
8D10 Initially it may not appear that an expression such as – 4 + 9x2 is a difference of perfect squares.
However, swapping the position of the two terms makes – 4 + 9x2 = 9x2 – 4, which can be
factorised to (3x + 2)(3x – 2). Use this idea to factorise these difference of perfect squares.
–9 + x2a –121 + 16x2b –25a2 + 4c –y2 + x2d–25a2 + 4b2e –36a2b2 + c2f –16x2 + y2z2g –900a2 + b2h
11 Olivia factorises 16x2 – 4 to get (4x + 2)(4x – 2) but the answer says 4(2x + 1)(2x – 1).
a What should Olivia do to get from her answer to the actual answer?
b What should Olivia have done initially to avoid this issue?
12 Find and explain the error in this working and correct it.
9 – (x – 1)2 = (3 + x – 1)(3 – x – 1)
= (2 + x)(2 – x)
REAS
ONING
10–1210 10, 11
Factorising with fractions and powers of 4
13 Some expressions with fractions or powers of 4 can be factorised in a similar way. Factorise these.
x2 – 14
a x2 – 425
b 25x2 – 916
c x2
9– 1d
a2
4– b
2
9e 5x2
9– 54
f 7a2
25– 28b2
9g a2
8– b
2
18h
x4 – y4i 2a4 – 2b4j 21a4 – 21b4k x4
3– y
4
3l
ENRICH
MEN
T
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8E Factorisation by grouping
When an expression contains four terms, such as x2 + 2x – x – 2,
it may be possible to factorise it into a product of two binomial
terms like (x – 1)(x + 2). In such situations the method of
grouping is often used.
Let’s start: Two methods – same result
The four-term expression x2 – 3x – 3 + x is written on the board.
Factorising by grouping is a bit like arrangingscattered objects into some sort of order.
Tommy chooses to rearrange the terms to give x2 – 3x + x – 3 then factorises by grouping.
Sharon chooses to rearrange the terms to give x2 + x – 3x – 3 then also factorises by grouping.
• Complete Tommy and Sharon’s factorisation working.
Tommy
x2 – 3x + x – 3 = x(x – 3) + 1( )
= (x – 3)( )
Sharon
x2 + x – 3x – 3 = x( ) – 3( )
= (x + 1)( )
• Discuss the differences in the methods. Is there any difference in their answers?
• Whose method do you prefer?
Keyideas
Factorisation by grouping is a method which is often used to factorise a four-term expression.
• Terms are grouped into pairs and factorised separately.
• The common binomial factor is then taken out to complete the
factorisation.
• Terms can be rearranged to assist in the search of a common factor.
x2 + 3x – 2x – 6
= x(x + 3) – 2(x + 3)
= (x + 3)(x – 2)
Example 8 Factorising by grouping
Use the method of grouping to factorise these expressions.
x2 + 2x + 3x + 6a x2 + 3x – 5x – 15b
SOLUTION EXPLANATION
a x2 + 2x + 3x + 6 = (x2 + 2x) + (3x + 6)
= x(x + 2) + 3(x + 2)
= (x + 2)(x + 3)
Group the first and second pair of terms.
Factorise each group.
Take the common factor (x + 2) out of both
groups.
b x2 + 3x – 5x – 15 = (x2 + 3x) + (–5x – 15)
= x(x + 3) – 5(x + 3)
= (x + 3)(x – 5)
Group the first and second pair of terms.
Factorise each group.
Take out the common factor (x + 3).
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508 508 Chapter 8 Algebraic techniques
Example 9 Rearranging an expression to factorise by grouping
Factorise 2x2 – 9 – 18x + x using grouping.
SOLUTION EXPLANATION
2x2 – 9 – 18x + x = 2x2 + x – 18x – 9
= x(2x + 1) – 9(2x + 1)
= (2x + 1)(x – 9)
Rearrange so that each group has a common
factor.
Factorise each group then take out (2x + 1).
Alternatively:
2x2 – 9 – 18x + x = 2x2 – 18x + x – 9
= 2x(x – 9) + 1(x – 9)
= (x – 9)(2x + 1)
Alternatively, you can group in another order
where each group has a comman factor. Then
factorise.
The answer will be the same.
Exercise 8E
1 Expand each expression.
2(x – 1)a 3(a + 4)b –5(1 – a)c–2(3 – x)d a(a + 5)e b(2 – b)fx(x – 4)g y(4 – y)h x(a + 1) + 2(a + 1)ia(x – 3) + 5(x – 3)j b(x – 2) – 3(x – 2)k c(1 – x) – 4(1 – x)l
2 Copy and then fill in the missing information.
2(x + 1) + x(x + 1) = (x + 1)( )a 3(x + 3) – x(x + 3) = (x + 3)( )b5(x + 5) – x(x + 5) = (x + 5)( )c x(x + 7) + 4(x + 7) = (x + 7)( )da(x – 3) + (x – 3) = (x – 3)( )e a(x + 4) – (x + 4) = (x + 4)( )f(x – 3) – a(x – 3) = (x – 3)( )g (4 – x) + 2a(4 – x) = (4 – x)( )h
3 Take out the common binomial term to factorise each expression.
x(x – 3) – 2(x – 3)a x(x + 4) + 3(x + 4)b x(x – 7) + 4(x – 7)c3(2x + 1) – x(2x + 1)d 4(3x – 2) – x(3x – 2)e 2x(2x + 3) – 3(2x + 3)f3x(5 – x) + 2(5 – x)g 2(x + 1) – 3x(x + 1)h x(x – 2) + (x – 2)i
UNDE
RSTA
NDING
—1–3(½) 3(½)
4Example 8 Use the method of grouping to factorise these expressions.
x2 + 3x + 2x + 6a x2 + 4x + 3x + 12b x2 + 7x + 2x + 14cx2 – 6x + 4x – 24d x2 – 4x + 6x – 24e x2 – 3x + 10x – 30fx2 + 2x – 18x – 36g x2 + 3x – 14x – 42h x2 + 4x – 18x – 72ix2 – 2x – xa + 2aj x2 – 3x – 3xc + 9ck x2 – 5x – 3xa + 15al
5 Use the method of grouping to factorise these expressions. The HCF for each pair includes
a pronumeral.
3ab + 5bc + 3ad + 5cda 4ab – 7ac + 4bd – 7cdb 2xy – 8xz + 3wy – 12wzc5rs – 10r + st – 2td 4x2 + 12xy – 3x – 9ye 2ab – a2 – 2bc + acf
FLUE
NCY
4–6(½)4–5(½) 4–6(½)
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8E6 Factorise these expressions. Remember to use a factor of 1 where necessary, for example,
x2 – ax + x – a = x(x – a) + 1(x – a).
x2 – bx + x – ba x2 – cx + x – cb x2 + bx + x + bcx2 + cx – x – cd x2 + ax – x – ae x2 – bx – x + bf
FLUE
NCY
7Example 9 Factorise these expressions by first rearranging the terms.
2x2 – 7 – 14x + xa 5x + 2x + x2 + 10b 2x2 – 3 – x + 6xc3x – 8x – 6x2 + 4d 11x – 5a – 55 + axe 12y + 2x – 8xy – 3f6m – n + 3mn – 2g 15p – 8r – 5pr + 24h 16x – 3y – 8xy + 6i
8 What expanded expression factorises to the following?
(x – a)(x + 4)a (x – c)(x – d)b (x + y)(2 – z)c (x – 1)(a + b)d(3x – b)(c – b)e (2x – y)(y + z)f (3a + b)(2b + 5c)g (m – 2x)(3y + z)h
9 Note that x2 + 5x + 6 = x2 + 2x + 3x + 6 which can be factorised by grouping. Use a similar
method to factorise the following.
x2 + 7x + 10a x2 + 8x + 15b x2 + 10x + 24cx2 – x – 6d x2 + 4x – 12e x2 – 11x + 18f
PROB
LEM-SOLVING
7, 97 7, 8
10 xa – 21 + 7a – 3x could be rearranged in two different ways before factorising.
Method 1
xa + 7a – 3x – 21 = a(x + 7) – 3( )=
Method 2
xa – 3x + 7a – 21 = x(a – 3) + 7( )=
a Copy and complete both methods for the above expression.
b Use different arrangements of the four terms to complete the factorisation of the following
in two ways. Show working using both methods.
xb – 6 – 3b + 2xi xy – 8 + 2y – 4xii 4m2 – 15n + 6m – 10mniii2m + 3n – mn – 6iv 4a – 6b2 + 3b – 8abv 3ab – 4c – b + 12acvi
11 Make up at least three of your own four-term expressions that factorise to a binomial
product. Describe the method that you used to make up each four-term expression.
REAS
ONING
10, 1110 10
Grouping with more than four terms
12 Factorise by grouping.
2(a – 3) – x(a – 3) – c(a – 3)a b(2a + 1) + 5(2a + 1) – a(2a + 1)bx(a + 1) – 4(a + 1) – ba – bc 3(a – b) – b(a – b) – 2a2 + 2abdc(1 – a) – x + ax + 2 – 2ae a(x – 2) + 2bx – 4b – x + 2fa2 – 3ac – 2ab + 6bc + 3abc – 9bc2g 3x – 6xy – 5z + 10yz + y – 2y2h8z – 4y + 3x2 + xy – 12x – 2xzi –ab – 4cx + 3aby + 2abx + 2c – 6cyj
ENRICH
MEN
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Using a CAS calculator 8E: Expanding and factorisingThis activity is in the interactive textbook in the form of a printable PDF.
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510 510 Chapter 8 Algebraic techniques
8F Factorising quadratic trinomials EXTENDING
An expression that takes the form x2 + bx + c, where b and c are
constants, is an example of a monic quadratic trinomial which has
the coefficient of x2 equal to 1. To factorise a quadratic expression,
we need to use the distributive law in reverse. Consider the
expansion shown at right:
(x + 2)(x − 4) = x2 − 2x − 8
factorised form expanded formfactorising
expanding
If we examine the expansion above we can see how each term of the product is formed.
(x + 2)(x − 4) = x2 − 2x −8
Product of 2 and −4 is −8 (2 × (−4) = −8, the constant term)
Product of x and x is x2
(x + 2)(x − 4) = x2 − 2x − 8
x × (−4) = −4x
2 × x = 2x
Add −4x and 2x to give the middle term, −2x(−4 + 2 = −2, the coefficient of x)
Let’s start: So many choices
Mia says that since –2 × 3 = –6 then x2 + 5x – 6 must equal (x – 2)(x + 3).
• Expand (x – 2)(x + 3) to see if Mia is correct.
• What other pairs of numbers multiply to give –6?
• Which pair of numbers should Mia choose to correctly factorise x2 + 5x – 6?
• What advice can you give Mia when trying to factorise these types of trinomials?
Keyideas To factorise a quadratic trinomial of the form
x2 + bx + c, find two numbers which:
For example:
x2 – 3x – 10 = (x – 5)(x + 2) choose –5 and +2
since –5 × 2 = –10 and –5 + 2 = –3• multiply to give c and
• add to give b.
Check factorisation steps by expanding. check: (x – 5)(x + 2) = x2 + 2x – 5x – 10
= x2 – 3x – 10
Write the factors in any order. write (x – 5)(x + 2) or (x + 2)(x – 5)
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Number and Algebra 511 511
Example 10 Factorising quadratic trinomials
Factorise each of the following quadratic expressions.
x2 + 7x + 10a x2 + 2x – 8b x2 – 7x + 10c
SOLUTION EXPLANATION
a x2 + 7x + 10 = (x + 5)(x + 2) Factors of 10 include: (10, 1) and (5, 2).
The pair that adds to 7 is (5, 2).
b x2 + 2x – 8 = (x + 4)(x – 2) Factors of –8 are (–8, 1) or (8, –1) or (4, –2) or
(– 4, 2) and 4 + (–2) = 2 so choose (4, –2).
c x2 – 7x + 10 = (x – 2)(x – 5) Factors of 10 are: (10, 1) or (–10, –1) or (5, 2)
or (–5, –2).
To add to a negative (–7), both factors must
then be negative: –5 + (–2) = –7 so choose
(–5, –2).
Example 11 Factorising with a common factor
Factorise the quadratic expression 2x2 – 2x – 12.
SOLUTION EXPLANATION
2x2 – 2x – 12 = 2(x2 – x – 6)
= 2(x – 3)(x + 2)
First take out common factor of 2.
Factors of –6 are: (–6, 1) or (6, –1) or (–3, 2) or
(3, –2).
–3 + 2 = –1 so choose (–3, 2).
Exercise 8F
1 Expand these binomial products.
(x + 1)(x + 3)a (x + 2)(x + 7)b (x – 3)(x + 11)c(x – 5)(x + 6)d (x + 12)(x – 5)e (x + 13)(x – 4)f(x – 2)(x – 6)g (x – 20)(x – 11)h (x – 9)(x – 1)i
2 Decide what two numbers multiply to give the first number and add to give the second number.
6, 5a 10, 7b 12, 13c20, 9d –5, 4e –7, –6f–15, 2g –30, –1h 6, –5i18, –11j 40, –13k 100, –52l
UNDE
RSTA
NDING
—1–2(½) 2(½)
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512 512 Chapter 8 Algebraic techniques
8F3Example 10a Factorise each of the following quadratic expressions.
x2 + 3x + 2a x2 + 4x + 3b x2 + 8x + 12cx2 + 10x + 9d x2 + 8x + 7e x2 + 15x + 14fx2 + 6x + 8g x2 + 7x + 12h x2 + 10x + 16ix2 + 8x + 15j x2 + 9x + 20k x2 + 11x + 24l
4Example 10b Factorise each of the following quadratic expressions.
x2 + 3x – 4a x2 + x – 2b x2 + 4x – 5cx2 + 5x – 14d x2 + 2x – 15e x2 + 8x – 20fx2 + 3x – 18g x2 + 7x – 18h x2 + x – 12i
5Example 10c Factorise each of the following quadratic expressions.
x2 – 6x + 5a x2 – 2x + 1b x2 – 5x + 4cx2 – 9x + 8d x2 – 4x + 4e x2 – 8x + 12fx2 – 11x + 18g x2 – 10x + 21h x2 – 5x + 6i
6 Factorise each of the following quadratic expressions.
x2 – 7x – 8a x2 – 3x – 4b x2 – 5x – 6cx2 – 6x – 16d x2 – 2x – 24e x2 – 2x – 15fx2 – x – 12g x2 – 11x – 12h x2 – 4x – 12i
7Example 11 Factorise each of the following quadratic expressions by first taking out a common factor.
2x2 + 10x + 8a 2x2 + 22x + 20b 3x2 + 18x + 24c2x2 + 14x – 60d 2x2 – 14x – 36e 4x2 – 8x + 4f2x2 + 2x – 12g 6x2 – 30x – 36h 5x2 – 30x + 40i3x2 – 33x + 90j 2x2 – 6x – 20k 3x2 – 3x – 36l
FLUE
NCY
3–7(½)3–6(½) 3–7(½)
8 Find the missing term in these trinomials if they are to factorise using integers. For example: the
missing term in x2 + + 10 could be 7x because x2 + 7x + 10 factorises to (x + 5)(x + 2) and 5
and 2 are integers. There may be more than one answer in each case.
x2 + + 5a x2 – + 9b x2 – – 12c x2 + – 12d
x2 + + 18e x2 – + 18f x2 – – 16g x2 + – 25h
9 A backyard, rectangular in area, has a length 2 metres more than its width (x metres). Inside the
rectangle are three square paved areas each of area 5 m2 as shown. The remaining area is lawn.
a Find an expression for:
the total backyard areaithe area of lawn in expanded formiithe area of lawn in factorised form.iii
b Find the area of lawn if:
x = 10ix = 7.ii
(x + 2) metres
x metres
PROB
LEM-SOLVING
8(½), 98 8(½), 9
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Number and Algebra 513 513
8F10 The expression x2 – 6x + 9 factorises to (x – 3)(x – 3) = (x – 3)2, which is a perfect square.
Factorise these perfect squares.
x2 + 8x + 16a x2 + 10x + 25b x2 + 30x + 225cx2 – 2x + 1d x2 – 14x + 49e x2 – 26x + 169f2x2 + 4x + 2g 5x2 – 30x + 45h –3x2 + 36x – 108i
11 Sometimes it is not possible to factorise quadratic trinomials using integers. Decide which of the
following cannot be factorised using integers.
x2 – x – 56a x2 + 5x – 4b x2 + 7x – 6cx2 + 3x – 108d x2 + 3x – 1e x2 + 12x – 53f
REAS
ONING
10(½), 1110(½) 10(½)
Completing the square
12 It is useful to be able to write a simple quadratic trinomial in the
form (x + b)2 + c. This involves adding (and subtracting) a special
number to form the first perfect square. This procedure is called
completing the square. Here is an example.
x2 − 6x − 8 = x2 − 6x + 9 − 9 − 8
= (x − 3)(x − 3) − 17 = (x − 3)2 − 17
62
(− )2 = 9
Complete the square for these trinomials.
x2 – 2x – 8a x2 + 4x – 1b x2 + 10x + 3cx2 – 16x – 3d x2 + 18x + 7e x2 – 32x – 11f
ENRICH
MEN
T
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514 514 Chapter 8 Algebraic techniques
8G Factorising trinomials of the form ax 2 + bx+ c EXTENDING
So far we have factorised quadratic trinomials where the coefficient of x2 is 1, such as x2 – 3x – 40.
These are called monic trinomials. We will now consider non-monic trinomials where the coefficient of
x is not equal to 1 and is also not a common factor to all three terms, such as in 6x2 + x – 15. The method
used in this section uses grouping which was discussed in section 8E.
Let’s start: How the grouping method works
Consider the trinomial 2x2 + 9x + 10.
• First write 2x2 + 9x + 10 = 2x2 + 4x + 5x + 10 then factorise
by grouping.
• Note that 9x was split to give 4x + 5x and the product of
2x2 and 10 is 20x2. Describe the link between the pair of
numbers {4, 5} and the pair of numbers {2, 10}.
• Why was 9x split to give 4x + 5x and not, say, 3x + 6x?
• Describe how the 13x should be split in 2x2 + 13x + 15 so
it can be factorised by grouping.
• Now try your method for 2x2 – 7x – 15.
Keyideas
To factorise a trinomial of the form ax2 + bx + c by grouping, find two numbers which sum to give
b and multiply to give a× c.
For example:
5x2 + 13x – 6
= 5x2 + 15x – 2x – 6
= 5x(x + 3) – 2(x + 3)
= (x + 3)(5x – 2)
a× c = 5 × (–6) = –30 so the two numbers are 15
and –2 since 15 + (–2) = 13 and 15 × (–2) = –30.
Mentally check your factors by expanding your answer.
(x + 3)(5x − 2)
5x2 −6
15x−2x 15x− 2x= 13x
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Number and Algebra 515 515
Example 12 Factorising trinomials of the form ax 2 + bx+ c
Factorise 2x2 + 7x + 3.
SOLUTION EXPLANATION
2x2 + 7x + 3 = 2x2 + x + 6x + 3
= x(2x + 1) + 3(2x + 1)
= (2x + 1)(x + 3)
a× c = 2 × 3 = 6 then ask what factors of this
number (6) add to 7. The answer is 1 and 6, so
split 7x = x + 6x. Then factorise by grouping.
Example 13 Factorising trinomials with negative numbers
Factorise the quadratic trinomials.
10x2 + 9x – 9a 6x2 – 17x + 12b
SOLUTION EXPLANATION
a 10x2 + 9x – 9 = 10x2 + 15x – 6x – 9
= 5x(2x + 3) – 3(2x + 3)
= (2x + 3)(5x – 3)
10 × (–9) = –90 so ask what factors of –90 add
to give 9. Choose 15 and –6. Then complete
the factorisation by grouping.
b 6x2 – 17x + 12 = 6x2 – 9x – 8x + 12
= 3x(2x – 3) – 4(2x – 3)
= (2x – 3)(3x – 4)
6 × 12 = 72 so ask what factors of 72 add to
give –17. Choose –9 and –8.
Complete a mental check.
(2x − 3)(3x − 4)−9x−8x −9x− 8x= −17x
Exercise 8G
1 List the two numbers which satisfy each part.
Multiply to give 6 and add to give 5a Multiply to give 12 and add to give 8bMultiply to give –10 and add to give 3c Multiply to give –24 and add to give 5dMultiply to give 18 and add to give –9e Multiply to give 35 and add to give –12fMultiply to give –30 and add to give –7g Multiply to give –28 and add to give –3h
UNDE
RSTA
NDING
—1, 2(½) 2(½)
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516 516 Chapter 8 Algebraic techniques
8G2 Copy and complete.
2x2 + 7x + 5 = 2x2 + 2x + + 5
= 2x( ) + 5( )
= ( )( )
a 3x2 + 8x + 4 = 3x2 + 6x + + 4
= 3x( ) + 2( )
= ( )( )
b
2x2 – 7x + 6 = 2x2 – 3x – + 6
= x( ) – 2( )
= ( )( )
c 5x2 + 9x – 2 = 5x2 + 10x – – 2
= 5x( ) – 1( )
= ( )( )
d
4x2 + 11x + 6 = 4x2 + + 3x + 6
= (x + 2) + 3( )
= ( )( )
e 6x2 – 7x – 3 = 6x2 – 9x + – 3
= ( ) + 1( )
= ( )( )
f
UNDE
RSTA
NDING
3Example 12 Factorise these quadratic trinomials.
2x2 + 9x + 4a 3x2 + 7x + 2b 2x2 + 7x + 6c3x2 + 8x + 4d 5x2 + 12x + 4e 2x2 + 11x + 12f6x2 + 13x + 5g 4x2 + 5x + 1h 8x2 + 14x + 5i
4Example 13 Factorise these quadratic trinomials.
3x2 + 2x – 5a 5x2 + 6x – 8b 8x2 + 10x – 3c6x2 – 13x – 8d 10x2 – 3x – 4e 5x2 – 11x – 12f4x2 – 16x + 15g 2x2 – 15x + 18h 6x2 – 19x + 10i12x2 – 13x – 4j 4x2 – 12x + 9k 7x2 + 18x – 9l9x2 + 44x – 5m 3x2 – 14x + 16n 4x2 – 4x – 15o
FLUE
NCY
3–4(½)3–4(½) 3–4(½)
5 Factorise these quadratic trinomials.
10x2 + 27x + 11a 15x2 + 14x – 8b 20x2 – 36x + 9c18x2 – x – 5d 25x2 + 5x – 12e 32x2 – 12x – 5f27x2 + 6x – 8g 33x2 + 41x + 10h 54x2 – 39x – 5i12x2 – 32x + 21j 75x2 – 43x + 6k 90x2 + 33x – 8l
6 Factorise by firstly taking out a common factor.
30x2 – 14x – 4a 12x2 + 18x – 30b 27x2 – 54x + 15c21x2 – 77x + 42d 36x2 + 36x – 40e 50x2 – 35x – 60f
7 Factorise these trinomials.
–2x2 + 7x – 6a –5x2 – 3x + 8b –6x2 + 13x + 8c18 – 9x – 5x2d 16x – 4x2 – 15e 14x – 8x2 – 5f
PROB
LEM-SOLVING
5–6(½), 75(½) 5–6(½)
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Number and Algebra 517 517
8G8 When splitting the 3x in 2x2 + 3x – 20, you could write:
A 2x2 + 8x – 5x – 20 or B 2x2 – 5x + 8x – 20
Complete the factorisation using A.aComplete the factorisation using B.b
c Does the order matter when you split the 3x?
d Factorise these trinomials twice each. Factorise once by grouping then repeat but reverse
the order of the two middle terms in the first line of working.
3x2 + 5x – 12i 5x2 – 3x – 14ii 6x2 + 5x – 4iii
9 Make up five non-monic trinomials with the coefficient of x2 not equal to 1 which factorise using
the above method. Explain your method in finding these trinomials.
REAS
ONING
8, 98 8
The cross method
10 The cross method is another way to factorise trinomials of the form ax2 + bx + c. It involves
finding factors of ax2 and factors of c then choosing pairs of these factors that add to bx.
For example: Factorise 6x2 – x – 15.
Factors of 6x2 include (x, 6x) and (2x, 3x).
Factors of –15 include (15, –1), (–15, 1), (5, –3) and (–5, 3).
We arrange a chosen pair of factors vertically then cross-multiply and add to get –1x.
x × (−1) + 6x × 15 x × (−3) + 6x × 5 2x × (−5) + 3x × 3= −1x= 27x ≠ −1x= 89x ≠ −1x
15
−16x
x 5
. . . . . . . . . . .−36x
x 3
−53x (3x − 5)
(2x + 3)2x
You will need to continue until a particular combination works. The third cross-product gives a
sum of –1x so choose the factors (2x + 3) and (3x – 5) so:
6x2 – x – 15 = (2x + 3)(3x – 5)
Try this method on the trinomials from Questions 4 and 5.
ENRICH
MEN
T
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518 518 Chapter 8 Algebraic techniques
Progress quiz
138pt8A Expand the following.
(x + 4)(x + 2)a (a – 5)(a + 8)b(2x – 3)(x + 6)c (3a – b)(a – 2b)d
238pt8B Expand each of the following.
(y + 4)2a (x – 3)2b(2a – 3)2c (7k + 2m)2d
338pt8B Expand and simplify the following.
(x + 5)(x – 5)a (11x – 9y)(11x + 9y)b
438pt8C Factorise the following.
25a – 15a x2 – 7xb–12x2 – 16xc 2(a + 3) + a(a + 3)d7(8 + a) – a(8 + a)e k(k – 4) – (k – 4)f
538pt8D Factorise each of the following.
x2 – 81a 16a2 – 49b25x2 – y2c 2a2 – 50d12x2y2 – 12e (h + 3)2 – 64f
638pt8E Use the method of grouping to factorise these expressions.
x2 + 7x + 2x + 14a a2 + 5a – 4a – 20b x2 – hx + x – hc
738pt8E Use grouping to factorise these expressions by first rearranging.
2x2 – 9 – 6x + 3xa 3ap – 10 + 2p – 15ab
838pt8F Factorise each of the following quadratic expressions.
Extx2 + 6x + 8a a2 + 2a – 15bm2 – 11m + 30c 2k2 + 2k – 24d
938pt8G Factorise these quadratic trinomials.
Ext2k2 + 7k + 6a 2x2 + 11x + 5b3a2 + 10a – 8c 10m2 – 19m + 6d
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Number and Algebra 519 519
8H Simplifying algebraic fractions: multiplication and division
With a numerical fraction such as 69, the highest common factor of 6 and 9 is 3, which can be
cancelled 69=
1�3 × 21�3 × 3
= 23. For algebraic fractions the process is the same. If expressions are in a
factorised form, common factors can be easily identified and cancelled.
Let’s start: Correct cancelling
Consider this cancelling attempt:
5x + ��101
��202= 5x + 1
2
• Substitute x = 6 into the left-hand side to evaluate 5x + 1020
.
• Substitute x = 6 into the right-hand side to evaluate 5x + 12
.
• What do you notice about the two answers to the above? How can you explain this?
• Decide how you might correctly cancel the expression on the left-hand side. Show your steps and check
by substituting a value for x.
Keyideas
Simplify algebraic fractions by factorising and cancelling only common factors.
Incorrect
2x + �4 2
�21= 2x + 2
Correct
2x + 42
=1�2(x + 2)
�21
= x + 2
To multiply algebraic fractions:
• factorise expressions where possible
• cancel if possible
• multiply the numerators and the denominators.
To divide algebraic fractions:
• multiply by the reciprocal of the fraction following the division sign
• follow the rules for multiplication after converting to the reciprocal
- The reciprocal of abis ba.
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520 520 Chapter 8 Algebraic techniques
Example 14 Simplifying algebraic fractions
Simplify the following by cancelling.
3(x + 2)(x – 4)6(x – 4)
a 20 – 5x8 – 2x
b x2 – 4x + 2
c
SOLUTION EXPLANATION
a1�3(x + 2)���(x – 4)1
2�6���(x – 4)1= x + 2
2Cancel the common factors (x – 4) and 3.
b 20 – 5x8 – 2x
= 5���(4 – x)1
2���(4 – x)1
= 52
Factorise the numerator and denominator then
cancel common factor of (4 – x).
c x2 – 4x + 2
=1���(x + 2)(x – 2)
���(x + 2)1
= x – 2
Factorise the difference of squares in the
numerator then cancel the common factor.
Example 15 Multiplying and dividing algebraic fractions
Simplify the following.
3(x – 1)(x + 2)
× 4(x + 2)9(x – 1)(x – 7)
a (x – 3)(x + 4)x(x + 7)
÷ 3(x + 4)x + 7
bx2 – 425
×5x + 5
x2 – x – 2 Extc
SOLUTION EXPLANATION
a1�3���(x – 1)1
���(x + 2)1× 4���(x + 2)1
3�9���(x – 1)1(x – 7)
= 1 × 41 × 3(x – 7)
= 43(x – 7)
First, cancel any factors in the numerators with
a common factor in the denominators. Then
multiply the numerators and the denominators.
b (x – 3)(x + 4)x(x + 7)
÷ 3(x + 4)x + 7
= (x – 3)���(x + 4)1
x���(x + 7)1× ���(x + 7)1
3���(x + 4)1
= x – 33x
Multiply by the reciprocal of the fraction after the
division sign.
Cancel common factors and multiply remaining
numerators and denominators.
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Number and Algebra 521 521
c x2 – 425
×5x + 5
x2 – x – 2
=1���(x – 2)(x + 2)
��25 5 × �51���(x + 1)1
1���(x – 2)���(x + 1)1
= x + 25
First factorise all the algebraic expressions. Note
that x2 – 4 is a difference of perfect squares. Then
cancel as normal.
Exercise 8H
1 Simplify these fractions by cancelling.
515
a 2416
b 5x10
c 42x12
d
248x
e 918x
f 3(x + 1)6
g 22(x – 4)11
h
2 Factorise these by taking out common factors.
3x + 6a 20 – 40xb x2 – 7xc 6x2 + 24xd
3 Copy and complete.
2x – 48
= 2( )8
=4
a 12 – 18x2x – 3x2
= 6( )x( )
= 6x
b
x – 1x + 3
÷ 2(x – 1)(x + 3)(x + 2)
= x – 1x + 3
×
= x + 22
c
UNDE
RSTA
NDING
—1–3 3
4Example 14a Simplify the following by cancelling.
3(x + 2)4(x + 2)
a x(x – 3)3x(x – 3)
b 20(x + 7)5(x + 7)
c
(x + 5)(x – 5)(x + 5)
d 6(x – 1)(x + 3)9(x + 3)
e 8(x – 2)4(x – 2)(x + 4)
f
5Example 14b Simplify the following by factorising and then cancelling.
5x – 55
a 4x – 1210
b 2x – 43x – 6
c 12 – 4x6 – 2x
d
x2 – 3xx
e 4x2 + 10x5x
f 3x + 3y2x + 2y
g 4x – 8y3x – 6y
h
FLUE
NCY
5–9(½)4–8(½) 4–9(½)
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522 522 Chapter 8 Algebraic techniques
8H6Example 14c Simplify the following. These expressions involve difference of perfect squares.
x2 – 100x + 10
a x2 – 49x + 7
b x2 – 25x + 5
c
2(x – 20)
x2 – 400d
5(x – 6)
x2 – 36e
3x + 27
x2 – 81f
7Example 15a Simplify the following by cancelling.
2x(x – 4)4(x + 1)
× (x + 1)x
a (x + 2)(x – 3)x – 5
× x – 5x + 2
b
x – 3x + 2
× 3(x + 4)(x + 2)x + 4
c 2(x + 3)(x + 4)(x + 1)(x – 5)
× (x + 1)4(x + 3)
d
8Example 15b Simplify the following by cancelling.
x(x + 1)x + 3
÷ x + 1x + 3
a x + 3x + 2
÷ x + 32(x – 2)
b
x – 4(x + 3)(x + 1)
÷ x – 44(x + 3)
c 4xx + 2
÷ 8xx – 2
d
3(4x – 9)(x + 2)2(x + 6)
÷ 9(x + 4)(4x – 9)4(x + 2)(x + 6)
e 5(2x – 3)(x + 7)
÷ (x + 2)(2x – 3)x + 7
f
9Ext Simplify by firstly factorising.
x2 – x – 6x – 3
a x2 + 8x + 16x + 4
b x2 – 7x + 12x – 4
c
x – 2
x2 + x – 6d
x + 7
x2 + 5x – 14e
x – 9
x2 – 19x + 90f
FLUE
NCY
10Example 15c
Ext
These expressions involve a combination of trinomials, difference of perfect squares and simple
common factors. Simplify by firstly factorising where possible.
x2 + 5x + 6x + 5
÷x + 3
x2 – 25a
x2 + 6x + 8
x2 – 9÷ x + 4x – 3
b
x2 + x – 12
x2 + 8x + 16×
x2 – 16
x2 – 8x + 16c
x2 + 12x + 35
x2 – 25×x2 – 10x + 25
x2 + 9x + 14d
9x2 – 3x
6x – 45x2e
x2 – 4x
3x – x2f
3x2 – 21x + 36
2x2 – 32×2x + 106x – 18
g2x2 – 18x + 40
x2 – x – 12×
3x + 15
4x2 – 100h
PROB
LEM-SOLVING
1010(½) 10
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Number and Algebra 523 523
8H
11 The expression 5 – 2x2x – 5
can be written in the form – 1(–5 + 2x)2x – 5
= – 1(2x – 5)2x – 5
, which can be
cancelled to –1.
Use this idea to simplify these algebraic fractions.
7 – 3x3x – 7
a 4x – 11 – 4x
b 8x + 16– 2 – x
c
x + 3– 9 – 3x
d 5 – 3x18x – 30
e x2 – 93 – x
f
12Ext Just like a2
2acan be cancelled to a
2, (a + 5)2
2(a + 5)cancels to a + 5
2. Use this idea to cancel these
fractions.
(a + 1)2
(a + 1)a 5(a – 3)2
(a – 3)b 7(x + 7)2
14(x + 7)c
3(x – 1)(x + 2)2
18(x – 1)(x + 2)d x2 + 6x + 9
2x + 6 Exte 11x – 22x2 – 4x + 4 Extf
REAS
ONING
11, 1211(½) 11
All in together
13Ext Use your knowledge of factorisation and the ideas in Questions 11 and 12 above to simplify
these algebraic fractions.
2x2 – 2x – 2416 – 4x
ax2 – 14x + 49
21 – 3xb
x – 16x + 64
64 – x2c
4 – x2
x2 + x – 6×
2x + 6x + 4x + 4
d
2x2 – 18
x2 – 6x + 9×
6 – 2x
x2 + 6x + 9e
x2 – 2x + 14 – 4x
÷1 – x2
3x2 + 6x + 3f
4x2 – 9
x2 – 5x÷
6 – 4x15 – 3x
gx2 – 4x + 4
8 – 4x×
–2
4 – x2h
(x + 2)2 – 4
(1 – x)2×x2 – 2x + 13x + 12
i2(x – 3)2 – 50
x2 – 11x + 24÷x2 – 43 – x
j
ENRICH
MEN
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524 524 Chapter 8 Algebraic techniques
8I Simplifying algebraic fractions: addition and subtractionThe process required for adding or subtracting
algebraic fractions is similar to that used for fractions
without pronumerals.
To simplify 23+ 45, for example, you would find the
lowest common multiple of the denominators (15)
then express each fraction using this denominator.
Adding the numerators completes the task.
Let’s start: Compare the working
Here is the working for the simplification of the sum
of a pair of numerical fractions and the sum of a pair
of algebraic fractions.
Although algebraic fractions, seem abstract, performingoperations on them and simplifying them is essentialto many calculations in real-life mathematical problems.
25+ 34= 820
+ 1520
= 2320
2x5
+ 3x4
= 8x20
+ 15x20
= 23x20
• What type of steps were taken to simplify the algebraic fractions that are the same as for the numerical
fractions?
• Write down the steps required to add (or subtract) algebraic fractions.
Keyideas
To add or subtract algebraic fractions:
• determine the lowest common denominator (LCD)
• express each fraction using the LCD
• add or subtract the numerators.
Example 16 Adding and subtracting with numerals in the denominators
Simplify:x4– 2x
5a 7x
3+ x6
b x + 32
+ x – 25
c
SOLUTION EXPLANATION
a x4– 2x
5= 5x20
– 8x20
= – 3x20
Determine the LCD of 4 and 5, i.e. 20. Express
each fraction as an equivalent fraction with a
denominator of 20. 2x× 4 = 8x. Then subtract
numerators.
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Number and Algebra 525 525
b 7x3
+ x6= 14x
6+ x6
= 15x6
= 5x2
Note the LCD of 3 and 6 is 6 not 3 × 6 = 18.
Simplify 156
to 52in the final step.
c x + 32
+ x – 25
= 5(x + 3)10
+ 2(x – 2)10
= 5x + 15 + 2x – 410
= 7x + 1110
The LCD of 2 and 5 is 10, write as equivalent
fractions with denominator 10.
Expand the brackets and simplify the numerator
by adding and collecting like terms.
Example 17 Adding and subtracting with algebraic terms in the denominators
Simplify:
2x–
52x
a2x+
3
x2b
SOLUTION EXPLANATION
a 2x– 52x
= 42x
– 52x
= – 12x
The LCD of x and 2x is 2x, so rewrite the
first fraction in an equivalent form with a
denominator also of 2x.
b2x+
3
x2=
2x
x2+
3
x2
=2x + 3
x2
The LCD of x and x2 is x2 so change the first
fraction so its denominator is also x2, then add
numerators.
Exercise 8I
1 Find the lowest common multiple of these pairs of numbers.
(6, 8)a (3, 5)b (11, 13)c (12, 18)d
2 Write equivalent fractions by stating the missing expression.
2x5
=10
a 7x3
=9
b x + 14
=(x + 1)12
c
3x + 511
=(3x + 5)22
d 4x=
2xe 30
x + 1=
3(x + 1)f
UNDE
RSTA
NDING
—1–4 3
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526 526 Chapter 8 Algebraic techniques
8I3 Copy and complete these simplifications.
x4+ 2x
3=
12+
12=
12a 5x
7– 2x
5=
35–
35=
35b
x + 12
+ 2x + 34
=(x + 1)4
+2x + 3
4=
+ 2x + 3
4=
4c
4 Write down the LCD for these pairs of fractions.
x3, 2x5
a 3x7
, x2
b – 5x4
, x8
c 2x3
, – 5x6
d 7x10
, – 3x5
e
UNDE
RSTA
NDING
5Example 16a,b Simplify:
x7+ x2
a x3+ x15
b x4– x8
c x9+ x5
d
y7– y8
e a2+ a11
f b3– b9
g m3– m6
h
m6+ 3m
4i a
4+ 2a
7j 2x
5+ x10
k p9– 3p
7l
b2– 7b
9m 9y
8+ 2y
5n 4x
7– x5
o 3x4
– x3
p
6Example 16c Simplify:
x + 12
+ x + 35
a x + 33
+ x – 44
b a – 27
+ a – 58
c
y + 45
+ y – 36
d m – 48
+ m + 65
e x – 212
+ x – 38
f
2b – 36
+ b + 28
g 3x + 86
+ 2x – 43
h 2y – 57
+ 3y + 214
i
2t – 18
+ t – 216
j 4 – x3
+ 2 – x7
k 2m – 14
+ m – 36
l
7Example 17a Simplify:
3x+ 52x
a 73x
– 2x
b 74x
– 52x
c 43x
+ 29x
d
34x
– 25x
e 23x
+ 15x
f – 34x
– 7x
g – 53x
– 34x
h
FLUE
NCY
5–7(½)5–6(½) 5–7(½)
8Example 17b Simplify:3x+
2
x2a
5
x2+
4x
b7x+
3
x2c
4x–
5
x2d
3
x2–8x
e –4
x2+
1x
f3x–
7
2x2g –
23x
+3
x2h
PROB
LEM-SOLVING
8–10(½)8(½) 8–9(½)
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Number and Algebra 527 527
8I9 Simplify these mixed algebraic fractions.
2x+ x4
a – 5x
+ x2
b – 2x
– 4x3
c 32x
– 5x4
d
3x4
– 56x
e 13x
– x9
f – 25x
+ 3x2
g – 54x
– 3x10
h
10 Find the missing algebraic fraction. The fraction should be in simplest form.
x2+ = 5x
6a + x
4= 3x
8b 2x
5+ = 9x
10c
2x3
– = 7x15
d – x3= 5x
9e 2x
3– = 5x
12f
PROB
LEM-SOLVING
11 Find and describe the error in each set of working. Then find the correct answer.
4x5
– x3= 3x
2a x + 1
5+ x2= 2x + 1
10+ 5x10
= 7x + 110
b
5x3
+ x – 12
= 10x6
+ 3x – 16
= 13x – 16
c2x–
3
x2=
2
x2–
3
x2
=–1
x2
d
12 A student thinks that the LCD to use when simplifying x + 12
+ 2x – 14
is 8.
a Complete the simplification using a common denominator of 8.
b Now complete the simplification using the actual LCD of 4.
c How does your working for parts a and b compare? Which method is preferable and why?RE
ASON
ING
11, 1211 11, 12
More than two fractions!
13 Simplify by finding the LCD.2x5
– 3x2
– x3
a x4– 2x
3+ 5x
6b 5x
8– 5x
6+ 3x
4c
x + 14
+ 2x – 13
– x5
d 2x – 13
– 2x7
+ x – 36
e 1 – 2x5
– 3x8
+ 3x + 12
f
23x
+ 5x– 1x
g – 12x
+ 2x– 43x
h – 45x
– 12x
+ 34x
i
4
x2+ 32x
– 53x
j 5x–
3
2x2– 57x
k2
x2– 49x
–5
3x2l
2x+ x5– x3
m 3x2
– 12x
+ x3
n – 4x9
+ 25x
+ 2x5
o
ENRICH
MEN
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528 528 Chapter 8 Algebraic techniques
8J Further simplification of algebraic fractions EXTENDING
More complex addition and subtraction of algebraic fractions involves expressions like:
2x – 13
– x + 44
and 2x – 3
–5
(x – 3)2
In such examples, care needs to be taken at each step in the working to avoid common errors.
Let’s start: Three critical errors
The following simplification of algebraic fractions has three
critical errors. Can you find them?
2x + 13
– x + 22
= 2x + 16
– 3(x + 2)6
= 2x + 1 – 3x + 66
= x + 76
The correct answer is x – 46
.
Fix the solution to produce the correct answer.
Keyideas
When combining algebraic fractions which involve subtraction signs, recall that:
• the product of two numbers of opposite sign is a negative number
• the product of two negative numbers is a positive number.
For example:2(x – 1)
6–3(x + 2)
6=
2x – 2 – 3x –6
6
and5(1 – x)
8–2(x – 1)
8=
5 – 5x – 2x +2
8
A common denominator can be a product of two binomial terms.
For example: 2x + 3
+ 3x – 1
= 2(x – 1)(x + 3)(x – 1)
+ 3(x + 3)(x + 3)(x – 1)
= 2x – 2 + 3x + 9(x + 3)(x – 1)
= 5x + 7(x + 3)(x – 1)
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Number and Algebra 529 529
Example 18 Simplifying with more complex numerators
Simplify:
x – 13
– x + 45
a 2x – 36
– 3 – x5
b
SOLUTION EXPLANATION
a x – 13
– x + 45
= 5(x – 1)15
– 3(x + 4)15
= 5x – 5 – 3x – 1215
= 2x – 1715
The LCD of 3 and 5 is 15. Insert brackets
around each numerator when multiplying.
Note: –3(x + 4) = –3x – 12 not –3x + 12.
b 2x – 36
– 3 – x5
= 5(2x – 3)30
– 6(3 – x)30
= 10x – 15 – 18 + 6x30
= 16x – 3330
Determine the LCD and express as equivalent
fractions. Insert brackets.
Expand the brackets, recall –6 × (–x) = 6x and
then simplify the numerator.
Example 19 Simplifying with more complex denominators
Simplify:4
x + 1+
3x – 2
a3
(x – 1)2–
2x – 1
b
SOLUTION EXPLANATION
a 4x + 1
+ 3x – 2
= 4(x – 2)(x + 1)(x – 2)
+ 3(x + 1)(x + 1)(x – 2)
= 4x – 8 + 3x + 3(x + 1)(x – 2)
= 7x – 5(x + 1)(x – 2)
The lowest common multiple of (x + 1) and
(x – 2) is (x + 1)(x – 2). Rewrite each fraction
as an equivalent fraction with this denominator
then add numerators.
b3
(x – 1)2–
2x – 1
=3
(x – 1)2–2(x – 1)
(x – 1)2
=3 – 2x + 2
(x – 1)2
=5 – 2x
(x – 1)2
Just like the LCD of 32 and 3 is 32, the LCD
of (x – 1)2 and x – 1 is (x – 1)2.
Remember that –2(x – 1) = –2x + 2.
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530 530 Chapter 8 Algebraic techniques
Exercise 8J
1 Expand the following.
–2(x + 3)a –5(x + 1)b –7(2 + 3x)c–3(x – 1)d –10(3 – 2x)e –16(1 – 4x)f
2 Write the LCD for these pairs of fractions.
13, 59
a 316
, 18
b 3x,
5
x2c – 5
2x, 32
d
3x – 1
, 2x + 1
e 7x – 2
, 3x + 3
f 42x – 1
, – 1x – 4
g5
(x + 1)2, 4x + 1
h
UNDE
RSTA
NDING
—1, 2 2(½)
3Example 18a Simplify:
x + 34
– x + 23
a x – 13
– x + 35
b x – 43
– x + 16
c
3 – x5
– x + 42
d 5x – 14
– 2 + x8
e 3x + 214
– x + 44
f
1 + 3x4
– 2x + 36
g 2 – x5
– 3x + 13
h 2x – 36
– 4 + x15
i
4Example 18b Simplify:
x + 53
– x – 12
a x – 45
– x – 67
b 3x – 74
– x – 12
c
5x – 97
– 2 – x3
d 3x + 24
– 5 – x10
e 9 – 4x6
– 2 – x8
f
4x + 33
– 5 – 2x9
g 2x – 14
– 1 – 3x14
h 3x – 28
– 4x – 37
i
5Example 19a Simplify:
3x – 1
+ 4x + 1
a 5x + 4
+ 2x – 3
b 3x – 2
+ 4x + 3
c
3x – 4
+ 2x + 7
d 7x + 2
– 3x + 3
e 3x + 4
– 2x – 6
f
– 1x + 5
+ 2x + 1
g – 2x – 3
– 4x – 2
h 3x – 5
– 5x – 6
i
FLUE
NCY
3–5(½)3–5(½) 3–5(½)
6Example 19b Simplify:4
(x + 1)2–
3x + 1
a2
(x + 3)2–
4x + 3
b3
x – 2+
4
(x – 2)2c
–2x – 5
+8
(x – 5)2d
–1x – 6
+3
(x – 6)2e
2
(x – 4)2–
3x – 4
f
5
(2x + 1)2+
22x + 1
g9
(3x + 2)2–
43x + 2
h4
(1 – 4x)2–
51 – 4x
i
PROB
LEM-SOLVING
6–7(½)6(½) 6–7(½)
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Number and Algebra 531 531
8J7 Simplify:
3x
(x – 1)2+
2x – 1
a3x + 23x
+712
b2x – 1
4+
2 – 3x10x
c
2xx – 5
–x
x + 1d
34 – x
–2xx – 1
e5x + 1
(x – 3)2+
xx – 3
f
3x – 7
(x – 2)2–
5x – 2
g–7x
2x + 1+
3xx + 2
hx
x + 1–
5x + 1
(x + 1)2i
PROB
LEM-SOLVING
8 One of the most common errors made when subtracting
algebraic fractions is hidden in this working shown on the
right:
a What is the error and in which step is it made?
b By correcting the error how does the answer change?
7x2
– x – 25
= 35x10
– 2(x – 2)10
= 35x – 2x – 410
= 33x – 410
9 Simplify:
1(x + 3)(x + 4)
+ 2(x + 4)(x + 5)
a 3(x + 1)(x + 2)
– 5(x + 1)(x + 4)
b
4(x – 1)(x – 3)
– 6(x – 1)(2 – x)
c 5x(x + 1)(x – 5)
– 2x – 5
d
3x – 4
+ 8x(x – 4)(3 – 2x)
e 3x(x + 4)(2x – 1)
– x(x + 4)(3x + 2)
f
10 Use the fact that a – b = –1(b – a) to help simplify these.
31 – x
– 2x – 1
a 4x5 – x
+ 3x – 5
b 27x – 3
– 73 – 7x
c
14 – 3x
+ 2x3x – 4
d – 3x5 – 3x
– 53x – 5
e 4x – 6
+ 46 – x
fRE
ASON
ING
9, 10(½)8 8, 9(½)
Factorise first
11 Factorising a denominator before further simplification is a useful step. Simplify these by
firstly factorising the denominators if possible.
3x + 2
+5
2x + 4a
73x – 3
–2
x – 1b
38x – 4
–5
1 – 2xc
4
x2 – 9–
3x + 3
d
52x + 4
+2
x2 – 4e
103x – 4
–7
9x2 – 16f
7
x2 + 7x + 12+
2
x2 – 2x – 15g
3
(x + 1)2 – 4–
2
x2 + 6x + 9h
3
x2 – 7x + 10–
210 – 5x
i1
x2 + x–
1
x2 – xj
ENRICH
MEN
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532 532 Chapter 8 Algebraic techniques
8K Equations with algebraic fractions EXTENDING
For equations with more than one fraction it is often best to try to simplify the equation by dealing with
all the denominators at once. This involves finding and multiplying both sides by the lowest common
denominator.
Let’s start: Why use the LCD?
For this equation follow each instruction.
x + 13
+ x4= 1
• Multiply every term in the equation by 3. What effect does this have on the fractions on the left-hand
side?
• Starting with the original equation, multiply every term in the equation by 4. What effect does this have
on the fractions on the left-hand side?
• Starting with the original equation, multiply every term in the equation by 12 and simplify.
Which instruction above does the best job in simplifying the algebraic fractions? Why?
Keyideas
For equations with more than one fraction multiply both sides by the lowest commondenominator (LCD).• Multiply every term on both sides, not just the fractions.
• Simplify the fractions and solve the equation using the methods learnt earlier.
Alternatively, express each fraction using the same denominator then simplify by adding or
subtracting and solve.
Example 20 Solving equations involving algebraic fractions
Solve each of the following equations.2x3
+ x2= 7a x – 2
5– x – 1
3= 1b
52x
– 43x
= 2c 3x + 1
= 2x + 4
d
SOLUTION EXPLANATION
a 2x3
+ x2= 7
2x�31
× �6 2 + x�21
× �6 3 = 7 × 6
4x + 3x = 42
7x = 42
x = 6
Multiply each term by the LCD (LCD of 3 and 2
is 6) and cancel.
Simplify and solve for x.
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Number and Algebra 533 533
OR 2x3
+ x2= 7
4x6
+ 3x6
= 7
7x6
= 7
7x = 42
x = 6
Alternatively, write each term on the left-hand
side using the LCD = 6.
Simplify by adding the numerators and solve the
remaining equation.
b x – 25
– x – 13
= 1
��15 3(x – 2)�51
– ��15 5(x – 1)�31
= 1 × 15
3(x – 2) – 5(x – 1) = 15
3x – 6 – 5x + 5 = 15
–2x – 1 = 15
–2x = 16
x = –8
Multiply each term on both sides by 15 (LCD of
3 and 5 is 15) and cancel.
Expand the brackets and simplify by combining
like terms. Note: –5(x – 1) = –5x + 5 not –5x – 5.
(Alternatively, write each term using the
LCD = 15 then combine the numerators and
solve. 3(x – 2)15
– 5(x – 1)15
= 1)
c 52x
– 43x
= 2
5��2x1
×��6x 3 – 4��3x1
×��6x2 = 2 × 6x
15 – 8 = 12x
7 = 12x
x = 712
LCD of 2x and 3x is 6x.
Multiply each term by 6x. Cancel and simplify.
Solve for x leaving the answer in fraction form.
(Alternative solution: 156x
– 86x
= 2)
d 3x + 1
= 2x + 4
3���(x + 1)(x + 4)���(x + 1)
= 2(x + 1)���(x + 4)���(x + 4)
3(x + 4) = 2(x + 1)
3x + 12 = 2x + 2
x + 12 = 2
OR x = –10
3x + 1
= 2x + 4
3(x + 4) = 2(x + 1)
3x + 12 = 2x + 2
x + 12 = 2
x = –10
Multiply each term by the common denominator
(x + 1)(x + 4).
Expand the brackets.
Subtract 2x from both sides to gather x terms on
one side then subtract 12 from both sides.
Since each side is a single fraction you can
‘cross-multiply’:3
x + 12
x + 4
This gives the same result as above.
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534 534 Chapter 8 Algebraic techniques
Exercise 8K
1 Write down the lowest common denominator of all the fractions in these equations.x3– 2x
5= 1a x
2– 3x
4= 3b x + 1
3– x6= 5c
2x – 17
+ 5x + 24
= 6d x3+ x2= 15
e x – 14
– 3x8
= 18
f
2 Simplify the fractions by cancelling.12x3
a 21x7
b 4(x + 3)2
c
5(2x + 5)5
d 15x5x
e –7(x + 1)(x + 2)7(x + 1)
f
36(x – 7)(x – 1)9(x – 7)
g 18(3 – 2x)(1 – x)9(3 – 2x)
h –8(2 – 3x)(2x – 1)–8(2 – 3x)
i
UNDE
RSTA
NDING
—1–2(½) 2(½)
3Example 20a Solve each of the following equations.x2+ x5= 7a x
2+ x3= 10b y
3+ y4= 14c
x2– 3x
5= –1d 5m
3– m2
= 1e 3a5
– a3= 2f
3x4
– 5x2
= 14g 8a3
– 2a5
= 34h 7b2
+ b4= 15i
4Example 20b Solve each of the following equations.x – 12
+ x + 23
= 11a b + 32
+ b – 43
= 1b n + 23
+ n – 22
= 1c
a + 15
– a + 16
= 2d x + 52
– x – 14
= 3e x + 32
– x + 13
= 2f
m + 43
– m – 44
= 3g 2a – 82
+ a + 76
= 1h 2y – 14
– y – 26
= –1i
5 Solve each of the following equations.x + 12
= x3
a x – 23
= x2
b n + 34
= n – 12
c
a + 23
= a + 12
d 3 + y2
= 2 – y3
e 2m + 44
= m + 63
f
6Example 20c Solve each of the following equations.34x
– 12x
= 4a 23x
– 12x
= 2b 42m
– 25m
= 3c
12x
– 14x
= 9d 12b
+ 1b= 2e 1
2y+ 13y
= 4f
13x
+ 12x
= 2g 23x
– 1x= 2h 7
2a– 23a
= 1i
FLUE
NCY
3–7(½)3–5(½) 3–7(½)
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Number and Algebra 535 535
8K7Example 20d Solve each of the following equations.
3x + 1
= 1x + 2
a 2x + 3
= 3x + 2
b 2x + 5
= 3x – 2
c
1x – 3
= 12x + 1
d 2x – 1
= 12x + 1
e 1x – 2
= 23x + 2
f
FLUE
NCY
8 Half of a number (x) plus one-third of twice the same number is equal to 4.
a Write an equation describing the situation.
b Solve the equation to find the number.
9 Use your combined knowledge of all the methods learnt earlier to solve these equations with
algebraic fractions.2x + 31 – x
= 4a 5x + 2x + 2
= 3b 3x – 2x – 1
= 2c
2x3
+ x – 14
= 2x – 1d3
x2– 2x= 5x
e1 – 3x
x2+ 32x
= 4x
f
x – 12
+ 3x – 24
= 2x3
g 4x + 13
– x – 36
= x + 56
h 1x + 2
– 2x – 3
= 5(x + 2)(x – 3)
i
10 Molly and Billy each have the same number
of computer games (x computer games each).
Hazel takes one-third of Molly’s computer
games and a quarter of Billy’s computer
games to give her a total of 77 computer
games.
a Write an equation describing the total
number of computer games for Hazel.
b Solve the equation to find how many
computer games Molly and Billy
each had.
PROB
LEM-SOLVING
9(½), 108 8, 9(½)
11 A common error when solving equations with
algebraic fractions is made in this working. Find the
error and explain how to avoid it.
3x – 14
+ 2x = x3
(LCD = 12)
12(3x – 1)4
+ 2x = 12x3
3(3x – 1) + 2x = 4x
9x – 3 + 2x = 4x
7x = 3
x = 37
REAS
ONING
12, 1311, 12 11, 12
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536 536 Chapter 8 Algebraic techniques
8K12 Another common error is made in this working. Find
and explain how to avoid this error.x2– 2x – 1
3= 1 (LCD = 6)
6x2
– 6(2x – 1)3
= 6
3x – 2(2x – 1) = 6
3x – 4x – 2 = 6
–x = 8
x = –8
13 Some equations with decimals can by solved by firstly multiplying by a power of 10. Here
is an example.
0.8x – 1.2 = 2.5
8x – 12 = 25 Multiply both sides by 10 to remove all decimals.
8x = 37
x = 378
Solve these decimal equations using the same idea. For parts d–f you will need to multiply
by 100.
0.4x + 1.4 = 3.2a 0.3x – 1.3 = 0.4b 0.5 – 0.2x = 0.2c1.31x – 1.8 = 2.13d 0.24x + 0.1 = 3.7e 2 – 3.25x = 8.5f
REAS
ONING
Literal equations
14 Solve each of the following equations for x in terms of the other pronumerals.
Hint: you may need to use factorisation.xa– x2a
= ba axb
– cx2
= db x – ab
= xc
c
x + ab
= d + ec
d ax + b4
= x + c3
e x + a3b
+ x – a2b
= 1f
2a – ba
+ ax= ag 1
a– 1x= 1c
h ax= bc
i
ax+ b = c
xj ax – b
x – b= ck cx + b
x + a= dl
2a + xa
= bm 1x – a
= 1ax + b
n ab– aa + x
= 1o
ENRICH
MEN
T
14— —
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Number and Algebra 537 537
InvestigationExpanding quadratics using areasConsider the expansion of the quadratic (x + 3)(x + 6). This can be represented by finding the area of the
rectangle shown.
Total area = A1 + A2 + A3 + A4
= x2 + 6x + 3x + 18
Therefore:
(x + 3)(x + 6) = x2 + 9x + 18A1
A3
A2
A4
x
x 6
3
Expanding with positive signs
a Draw a diagram and calculate the area to determine the expansion of the following quadratics.
(x + 4)(x + 5)i (x + 7)(x + 8)ii
(x + 3)2iii (x + 5)2iv
b Using the same technique establish the rule for expanding (a + b)2.
Expanding with negative signs
Consider the expansion of (x – 4)(x – 7).
Area required = total area – (A2 + A3 + A4)
= x2 – [(A2 + A4) + (A3 + A4) – A4]
= x2 – (7x + 4x – 28)
= x2 – 11x + 28
Therefore:
(x – 4)(x – 7) = x2 – 11x + 28
A1
A3
A2
A4
x
x
4
This area is counted twicewhen we add 7x + 4x.
7
a Draw a diagram and calculate the area to determine the expansion of the following quadratics.
(x – 3)(x – 5)i
(x – 6)(x – 4)ii
(x – 4)2iii
(x – 2)2iv
b Using the same technique, establish the rule for expanding (a – b)2.
Difference of perfect squares
Using a diagram to represent (a – b)(a + b), determine the appropriate area and establish a rule for the
expansion of (a – b)(a + b).
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538 538 Chapter 8 Algebraic techniques
Numerical applications of perfect squaresThe expansion and factorisation of perfect squares and difference of perfect squares can be applied to
the mental calculation of some numerical problems.
Evaluating a perfect square
The perfect square 322 can be evaluated using (a + b)2 = a2 + 2ab + b2.
322 = (30 + 2)2 (Let a = 30, b = 2)
= 302 + 2(30)(2) + 22
= 900 + 120 + 4
= 1024
a Use the same technique to evaluate these perfect squares.
222i 212ii 332iii 512iv
1.22v 3.22vi 6.12vii 9.012viii
Similarly, the perfect square 292 can be evaluated using (a – b)2 = a2 – 2ab + b2.
292 = (30 – 1)2 (Let a = 30, b = 1)
= 302 – 2(30)(1) + 12
= 900 – 60 + 1
= 841
b Use the same technique to evaluate these perfect squares.
192i 392ii 982iii 872iv
1.92v 4.72vi 8.82vii 3.962viii
Evaluating the difference of perfect squares
The difference of perfect squares 142 – 92 can be evaluated using a2 – b2 = (a + b)(a – b).
142 – 92 = (14 + 9)(14 – 9) (Let a = 14, b = 9)
= 23 × 5
= 115
a Use the same technique to evaluate these difference of perfect squares.
132 – 82i 252 – 232ii 422 – 412iii 852 – 832iv
1.42 – 1.32v 4.92 – 4.72vi 10012 – 10002vii 2.012 – 1.992viii
The expansion (a + b)(a – b) = a2 – b2 can also be used to evaluate some products. Here is
an example:
31 × 29 = (30 + 1)(30 – 1) (Let a = 30, b = 1)
= 302 – 12
= 900 – 1
= 899
b Use the same technique to evaluate these products.
21 × 19i 32 × 28ii 63 × 57iii 105 × 95iv
2.1 × 1.9v 7.4 × 6.6vi 520 × 480vii 915 × 885viii
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Number and Algebra 539 539
Problems and challengesUp for a challenge? If you get stuck on a question,
check out the 'Working with Unfamiliar Questions' poster
at the end of the book to help you.
1 a The difference between the squares of two consecutive numbers is 97.
What are the two numbers?
b The difference between the squares of two consecutive odd numbers is 136.
What are the two numbers?
c The difference between the squares of two consecutive multiples of 3 is 81.
What are the two numbers?
2 a If x2 + y2 = 6 and (x + y)2 = 36, find the value of xy.
b If x + y = 10 and xy = 2, find the value of 1x+ 1y.
3 Find the values of the different digits a, b, c and d if the four digit number abcd× 4 = dcba.
4 a Find the quadratic rule that relates the width n to the number of matches in the pattern below.
n = 1 n = 2 n = 3
b Draw a possible pattern for these rules.
n2 + 3i
n(n – 1)ii
5 Factorise n2 – 1 and use the factorised form to explain why when n is prime and greater than 3,
n2 – 1 is:
divisible by 4i
divisible by 3ii
thus divisible by 12.iii
6 Prove that this expression is equal to 1.
2x2 – 8
5x2 – 5÷ x – 2
5x – 5÷2x2 – 10x – 28
x2 – 6x – 7
7 Prove that 4x2 – 4x + 1 ≥ 0 for all x.
8 In a race over 4 km Ryan ran at a constant speed. Sophie, however, ran the first 2 km
at a speed 1 km/h more than Ryan and ran the second 2 km at a speed 1 km/h less
than Ryan. Who won the race?
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540 540 Chapter 8 Algebraic techniques
Solving equations with algebraic fractions (Ext)
Find lowest common denominator(LCD) and multiply every termby the LCD.e.g. LCD of 2x and 3x is 6x LCD of (x − 1) and (x + 3) is (x − 1)(x + 3) LCD of 2x and x2 is 2x2
Factorisation
The reverse process of expansion.Always remove the highest commonfactor first.e.g. 2ab + 8b HCF = 2b = 2b (a + 4)e.g. 3(x + 2) + y (x + 2) HCF = x + 2 = (x + 2)(3 + y )
By grouping
If there are four terms, we may be able to group into two binomial terms.e.g. 5x + 10 − 3x2 − 6x
Monic quadratic trinomials (Ext)
These are of the form:x2 + bx + c
Require two numbers that multiplyto c and add to be.g. x2 − 7x −18 = (x − 9)(x + 2)since −9 × 2 = −18and −9 + 2 = −7
Trinomials of the form ax 2 + bx + c (Ext)
Can be factorised using grouping alsoe.g. 3x 2 + 7x − 6, 3 × −6 = −18 = 3x 2 + 9x − 2x − 6, Use 9 and –2= 3x (x + 3) − 2(x + 3) since= (x + 3)(3x − 2) 9 × (−2) = −18 and 9 + (−2) = 7
Special cases
DOPS (difference of perfect squares)
(a b)(a + b) = a2 b2
e.g. (x −− −
−
−
−
−
5)(x + 5) = x2 + 5x 5x 25 = x2 − 25Perfect squares
(a + b)2 = a2 + 2ab + b2
(a b)2 = a2 2ab + b2
e.g. (2x + 3)2 = (2x + 3)(2x + 3) = 4x2 + 6x + 6x + 9 = 4x2 + 12x + 9
Expansion
The process of removing brackets.
e.g. 2(x + 5) = 2x + 10
(x + 3)(2x − 4) = x(2x − 4) + 3(2x − 4) = 2x 2 − 4x + 6x − 12 = 2x 2 + 2x − 12
2
5
x
Algebraic techniques
2xx − 7
35x
7(x − 1)2
Algebraic fractions
These involve algebraic expressionsin the numerator and/or denominator.
e.g. , ,
DOPS
÷
×=
=
Multiply/divide
To divide, multiply by the reciprocal.Factorise all expressions, canceland then multiply.
3x2 − 9
152x + 6
3(x − 3)(x + 3)
2(x + 3)15
e.g.
1
5
1
1
25(x − 3)
Add/subtract
Must find lowest commondenominator (LCD) beforeapplying operation.
=
3x4
9x + 4x − 412
= +9x 12
4(x − 1) 12
= 13x − 412
e.g. + x −13
a2 − b2 = (a – b)(a + b)e.g. x2 − 16 = x2 − 42
e.g. 4x2 − 9 = (2x )2 − 32
= (x – 4)(x + 4)
= (2x − 3)(2x + 3)
= 5(x + 2) − 3x (x + 2)= (x + 2)(5 − 3x )
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Number and Algebra 541 541
Multiple-choice questions
138pt8A (2x – 1)(x + 5) in expanded and simplified form is:
2x2 + 9x – 5A x2 + 11x – 5B 4x2 – 5C3x2 – 2x + 5D 2x2 + 4x – 5E
238pt8B (3a + 2b)2 is equivalent to:
9a2 + 6ab + 4b2A 9a2 + 4b2B 3a2 + 6ab + 2b2C3a2 + 12ab + 2b2D 9a2 + 12ab + 4b2E
338pt8D 16x2 – 49 in factorised form is:
(4x – 7)2A (16x – 49)(16x + 49)B (2x – 7)(8x + 7)C(4x – 7)(4x + 7)D 4(4x2 – 49)E
438pt8E The factorised form of x2 + 3x – 2x – 6 is:
(x – 3)(x + 2)A x – 2(x + 3)2B (x + 3)(x – 2)Cx – 2(x + 3)D x(x + 3) – 2E
538pt8F If (x – 2) is a factor of x2 + 5x – 14 the other factor is:
ExtxA x + 7B x – 7C x – 16D x + 5E
638pt8G The factorised form of 3x2 + 10x – 8 is:
AExt
(3x + 1)(x – 8)
B (x – 4)(3x + 2)
C (3x + 2)(x + 5)
D (3x – 2)(x + 4)
E (x + 1)(3x – 8)
738pt8H The simplified form of 3x + 6(x + 5)(x + 1)
× x + 5x + 2
is:
3x + 1
A15
2x2 + 5B x + 3C 5
x + 2D 3(x + 5)E
838pt8I x + 25
+ 2x – 13
written as a single fraction is:
11x + 115
A 11x + 98
B 3x + 18
C 11x + 715
D 13x + 115
E
938pt8J The LCD of 3x + 12x
and 4x + 1
is:
Ext 8xA 2x(x + 1)B (x + 1)(3x + 1)C8x(3x + 1)D x(x + 1)E
1038pt8K The solution to 31 – x
= 42x + 3
is:
Ext x = – 12
A x = – 911
B x = 17
C x = 2D x = – 45
E
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542 542 Chapter 8 Algebraic techniques
Short-answer questions
138pt8A Expand the following binomial products.
(x – 3)(x + 4)a (x – 7)(x – 2)b(2x – 3)(3x + 2)c 3(x – 1)(3x + 4)d
238pt8B Expand the following.
(x + 3)2a (x – 4)2b (3x – 2)2c(x – 5)(x + 5)d (7 – x)(7 + x)e (11x – 4)(11x + 4)f
338pt8C Write the following in fully factorised form by removing the highest common factor.
4a + 12ba 6x – 9x2b –5x2y – 10xyc3(x – 7) + x(x – 7)d x(2x + 1) – (2x + 1)e (x – 2)2 – 4(x – 2)f
438pt8D Factorise the following DOPS.
x2 – 100a 3x2 – 48b 25x2 – y2c49 – 9x2d (x – 3)2 – 81e 1 – x2f
538pt8E Factorise the following by grouping.
a x2 – 3x + 6x – 18
b 4x2 + 10x – 2x – 5
c 3x – 8b + 2bx – 12
638pt8F/G Factorise the following trinomials.
Extx2 + 8x + 15a x2 – 3x – 18b x2 – 7x + 6c 3x2 + 15x – 42d2x2 + 16x + 32e 5x2 + 17x + 6f 4x2 – 4x – 3g 6x2 – 17x + 12h
738pt8H Simplify the following.
3x + 123
a 2x – 163x – 24
b x2 – 95(x + 3)
c
838pt8H Simplify the following algebraic fractions by first factorising and cancelling where possible.
32x
× x6
a x(x – 4)8(x + 1)
× 4(x + 1)x
b x2 + 3x3x + 6
× x + 2x + 3
c
2x5x + 20
÷ xx + 4
d4x2 – 9
10x2÷10x – 15
xe Ext
x2 + 5x + 6x + 3
÷x2 – 44x – 8
f
938pt8I/J Simplify the following by first finding the lowest common denominator.
x4+ 2x
3a Ext
x – 16
– x + 38
b 34x
+ 12x
c
7x–
2
x2d Ext
3x + 1
+ 5x + 2
e Ext7
(x – 4)2– 2x – 4
f
1038pt8K Solve the following equations involving fractions.
Extx4+ 2x
5= 13a 4
x– 23x
= 20b
x + 32
+ x – 43
= 6c 41 – x
= 5x + 4
d
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Extended-response questions
1 A pig pen for a small farm is being redesigned. It is originally a square of side length x m.
a In the planning the length is initially kept as x m and the width altered such that the area of the
pen is (x2 + 3x) square metres. What is the new width?
b Instead, it is determined that the original length will be increased by 1 metre and the original
width will be decreased by 1 metre.
What effect does this have on the perimeter of the pig pen compared with the original size?iDetermine an expression for the new area of the pig pen in expanded form. How does this
compare to the original area?
ii
c The final set of dimensions requires an extra 8 m
of fencing to go around the pen compared with
the original pen. If the length of the pen has been
increased by 7 m, then the width of the pen must
decrease. Find:
the change that has been made to the width of
the pen
i
the new area enclosed by the peniiwhat happens when x = 3.iii
2 The security tower for a palace is on a small square
piece of land 20 m by 20 m with a moat of width
x metres the whole way around it as shown.
a State the area of the piece of land.
20 m
Land
x m
b i Give expressions for the length and the width of
the combined moat and land.
ii Find an expression, in expanded form, for the
entire area occupied by the moat and the land.
c If the tower occupies an area of (x + 10)2 m2, what fraction of the total area in part b ii is this?d Use your answers to parts a and b to give an expression for the area occupied by the moat alone,
in factorised form.
e Use trial and error to find the value of x such that the area of the moat alone is 500 m2.
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