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52

2

Algebra

What would you do if you had control of the traffic lights?

When 000 FIRE is called, there is an extensive network of equipment that needs to be activated and personnel that need to be alerted. Who makes sure that all the technology is in place so that an emergency can be handled as smoothly as possible?

Miles Brown works as a Communication and Technical Services Officer at the CFA (Country Fire Authority) and does exactly that.

“We’re in charge of all the maintenance and installation of the communication and signalling systems for the CFA, from repairing radios and telephone equipment, to when there’s a fire call; all the things that make the lights go on, sirens go on, doors open, and traffic lights change.

There’s a lot of rural travel, working outdoors and physical tasks. Sometimes you can be running cabling, spending a day up in a roof or crawling around under a building. It’s very hands on. That’s one of the reasons that I chose it.”

Miles uses algebra to calculate the resistance of electronic circuits and measure the signal strength of radios.

Name: Miles BrownJob: Communication Technical Services OfficerQualifications: Advanced Diploma of Electronic Engineering

Why learn this?

Algebra gives us a shorthand way to describe and generalise patterns that occur in the real world. Algebraic techniques allow us to manipulate algebraic expressions into different forms to compare these patterns. In this chapter, you will acquire skills so that you can solve more complex equations and use variables to model a greater variety of situations.

After completing this chapter you will be able to:

•

expand binomial products, expressing answers in simplest form

•

factorise quadratic trinomials

•

recognise and use special products to factorise expressions

•

rearrange formulas to make a different variable the subject

•

simplify algebraic expressions involving fractions.

AA Pearson Maths 10 SB-02.fm Page 52 Friday, May 25, 2012 8:59 AM

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Recall

Prepare for this chapter by attempting the following questions. If you have difficulty with a question, go to Pearson Places and download the Recall Worksheet from Pearson Reader.

1

How many terms are there in each expression?

(a)

x

2

+

6

x

+

9

(b)

16

x

2

−

y

2

(c)

xy

+

2

y

+

4

x

+

8

2

What is the coefficient of each of these terms?

(a)

4

x

(b)

-6

ab

(c)

y

(d)

-12

x

2

y

3

Simplify:

(a)

3

×

a

+

5

×

b

(b)

8

×

3

x

−

8

×

y

4

Expand and simplify if possible:

(a)

3

(

x

+

2

)

(b)

5

(

x

−

3

)

(c)

3

x

+

2

(

x

−

4

)

5

Expand and simplify if possible:

(a)

-2

(

x

+

3

)

(b)

-8

(

x

−

5

)

(c)

-

(

x

+

2

)

6

Simplify:

(a)

(

3

x

)

2

(b)

(

2

y

)

2

(c)

(

7

k

)

2

7

Simplify:

(a)

(

4

x

)

2

−

2

x

2

(b)

p

2

−

(

3

p

)

2

(c)

(

-

y

)

2

+

y

2

8

Rearrange to make

x

the subject.

(a)

y

=

2

x

+

3

(b)

2

a

=

(c)

s

=

9

Find two numbers that:

(a)

multiply to give 21 and add to give 10

(b)

multiply to give 24 and add to give 14

(c)

multiply to give 20 and add to give -9

(d)

multiply to give -18 and add to give 3

(e)

multiply to give -20 and add to give -8

(f)

multiply to give -40 and add to give -3.

10

Simplify the following.

(a) (b) (c)

11

Simplify the following.

(a) (b) (c)

algebraic fractions factorise by grouping in pairs Perfect Square

binomial product factorising quadratic expression

completing the square highest common factor (HCF) quadratic trinomial

Difference of Two Squares monic quadratic

Worksheet R2.1

Worksheet R2.2

Worksheet R2.3

Worksheet R2.4

Worksheet R2.5

Worksheet R2.6

Worksheet R2.7

Worksheet R2.8x 4b–c

-------------- 5x at2–3

-------------------

Worksheet R2.9

Worksheet R2.1023--- 1

4---+ 5

6--- 3

8---– a

3--- 4a

3------+

Worksheet R2.1138--- 4

9---× x

10------ 5

x---× 5

6--- 10

9------÷

Key Words


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54 PEARSON mathematics 10–10A

Expanding brackets

Expanding bracketsThe Distributive Law is used to expand brackets.

a(b + c) = ab + aca(b − c) = ab − ac

Expanding binomial productsA binomial product is a product that has two terms in each expression. (x + 3)(x + 7) is an example of a binomial product. Remember that a multiplication sign is not required between the factors to indicate a product. We can understand how to find the expansion of binomial products by considering the rectangle below.

If the length of a rectangle is (x + 7) units and the width is (x + 3) units then an expression for the area is:

A = lwA = (x + 7)(x + 3) units2

We can separate the x + 7 into a length of x units and a length of 7 units. We separate (x + 3) in a similar way.

We can add the areas of the four smaller rectangles and simplify like terms.

A = x2 + 7x + 3x + 21A = x2 + 10x + 21 units2

Worked Example 1Expand and simplify each of the following.

(a) 3(4x − 2) (b) x(x + 4) − 2(x − 1)

Thinking(a) Multiply all terms inside the brackets by

the term in front of the brackets.(a) 3(4x − 2)

= 12x − 6

(b) 1 Expand both sets of brackets by multiplying every term inside the brackets by the term in front of the brackets. Change all signs when multiplying by a negative number.

(b) x(x + 4) − 2(x − 1) = x2 + 4x − 2x + 2

2 Simplify by collecting like terms. = x2 + 2x + 2

1

x × x =x2 units2

3 × x =3x units2

x × 7 =7x units2

3 × 7 =21 units2

x units 7 units

3 units

x units

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We can equate the two expressions that represent the area of the rectangle.

(x + 7)(x + 3) = x2 + 10x + 21

This is called the extended distributive law.

To help us expand, we can use FOIL. This tells us how to multiply the terms:

First terms in each bracketOutside terms in each bracketInside terms in each bracketLast terms in each bracket

and then add them all together.

Expanding special binomial productsThere are some expansions that are helpful to expand by rule:

The Difference of Two Squares rule:

(a + b)(a − b) = a2 − ab + ba − b2

= a2 − b2

The Perfect Square rules:

(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

To expand binomial products, multiply each term in the second bracket by each term in the first bracket and add them together.

(a + b)(c + d) = ac + ad + bc + bd

Worked Example 2Expand and simplify (x + 5)(2x − 3)

Thinking1 Multiply each term in the second bracket

by each term in the first bracket.

(x + 5)(2x − 3)

(x + 5)(2x − 3)= 2x2 − 3x + 10x − 15

2 Simplify by collecting like terms. = 2x2 + 7x − 15

These rules are used to expand special binomial products:

(a − b)(a + b) = a2 − b2 (Difference of Two Squares)(a + b)2 = a2 + 2ab + b2 (Perfect Square)(a − b)2 = a2 − 2ab + b2 (Perfect Square)

FO F O I L

I L

2

FO

I L


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Expanding three factorsIf we have three or more factors, we first multiply any two factors together and simplify the result. We then repeat this process until the expression is fully expanded. Look for special product expansions to expand more easily.

Worked Example 3Expand and simplify:

(a) (x − 3)(x + 3) (b) (3x + 1)2

Thinking(a) Identify that it has the form of a special

product and apply the Difference of Two Squares rule.

(a) (x − 3)(x + 3) = (x)2 − (3)2

= x2 − 9

(b) Identify that it has the form of a special product and apply the Perfect Square rule.

(b) (3x + 1)2 = (3x)2 + 2 × 3x × 1 + (1)2

= 9x2 + 6x + 1

Worked Example 4Expand and simplify.

(a) x(x − 2)(x + 3) (b) (x + 3)(x + 1)(x − 4) (c) (x − 1)(x − 5)(x + 1)

Thinking(a) 1 Expand and simplify the binomial

product.(a) x(x − 2)(x + 3)

= x(x2 + 3x − 2x − 6)= x(x2 + x − 6)

2 Multiply by the single factor. = x3 + x2 − 6x

(b) 1 Look for special product expansions. As there are none, choose any two brackets to expand. (Here, we will expand the first two brackets first and simplify their product.)

(b) (x + 3)(x + 1)(x − 4)= (x2 + x + 3x + 3)(x − 4)= (x2 + 4x + 3)(x − 4)

2 Expand the two remaining brackets by multiplying each term in the second bracket by each term in the first.

= x3 − 4x2 + 4x2 − 16x + 3x − 12

3 Simplify the expansion. = x3 − 13x − 12

(c) 1 Identify that two factors have the form of a special product and use this to expand these factors. Rearrange the expansion if necessary. (Here, two factors multiply to give the difference of two squares (x − 1)(x + 1) = x2 − 1).)

(c) (x − 1)(x − 5)(x + 1)= (x − 1)(x + 1)(x − 5)= (x2 − 1)(x − 5)

2 Expand the two remaining brackets. = x3 − 5x2 − x + 5

3

4

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Using your CAS to check expansionsThe algebra component of your CAS makes checking algebraic expansions very easy.

You should already be able to work out by hand that the expansion of x(x − 4)(x + 6) is x3 + 2x2 − 24x. On your CAS you use the expand function to find the expansion.

Expand the following and check the expansions using your CAS.

1 3x(x − 2)(x + 7) 2 (x − 2)(x + 8)(x + 2) 3 (x + 1)(x − 4)(x + 11)

If you have expanded the expressions correctly you should see these answers on your screen.

1 3x3 + 15x2 − 42x 2 x3 + 8x2 − 4x − 32 3 x3 + 8x2 − 37x − 44

Expanding brackets

Fluency1 Expand and simplify each of the following.

(a) 4(x + 3) (b) 8(x + 2) (c) 12(x + 11)

(d) 2(6x − 1) (e) 7(2x − 5) (f) x(5x − 4)

(g) -2(x + 1) (h) -x(x − 7) (i) -x(3x − 6)

(j) -3x(x + 6) (k) -2x(x − 9) (l) -4x(x + 2)

(m) x(x + 2) + 3(x + 2) (n) x(2x + 1) + 2(2x + 1) (o) x(5x + 2) + 3(5x + 2)

(p) 2x(x − 3) − 5(x + 6) (q) 5x(2x + 5) − 4(3x − 8) (r) 4x(3x − 1) − 6(2x + 1)

Using the TI-Nspire CAS Using the ClassPad

From the home screen, open a new document. You don’t need to save this so select ‘No’ when asked to save and press enter twice.

Press > Algebra > Expand and then enter the expression, being careful to include the correct number of brackets and a multiplication sign between each set of brackets in the expression.

From the menu, select Main .Tap Interactive > Transformation > Expand. A dialogue box will appear. Enter the expression then tap OK.

NavigatorQ1 Column 1, Q2 Column 1, Q3 Column 1, Q4 Column 1, Q5, Q6 Column 1, Q7, Q9, Q11, Q13, Q14, Q17

Q1 Column 2, Q2 Column 2, Q3 Column 2, Q4 Column 2, Q5, Q6 Column 2, Q7, Q8, Q10, Q12, Q13, Q14, Q15, Q18

Q1 Column 3, Q2 Column 3, Q3 Column 3, Q4 Column 3, Q5, Q6 Column 3, Q7, Q8, Q9, Q10, Q11, Q12, Q14, Q15, Q16, Q18, Q19

2.1Answerspage 719

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2 Expand and simplify each of the following.

(a) (x + 1)(x + 3) (b) (x + 4)(x + 2) (c) (x + 5)(x + 4)

(d) (x + 8)(x − 3) (e) (x + 7)(x − 9) (f) (x − 3)(x + 7)

(g) (x − 1)(x − 7) (h) (x − 4)(x − 5) (i) (x − 9)(x − 3)

(j) (2x + 3)(x − 2) (k) (4x − 5)(x + 3) (l) (7x + 1)(x − 4)

(m) (3x − 1)(2x − 5) (n) (5x − 3)(7 − x) (o) (6x − 5)(2 − 3x)

3 Expand and simplify each of the following, using the rules for special products.

(a) (2 − x)(2 + x) (b) (7 − x)(7 + x) (c) (6 + x)(6 − x)

(d) (2x + 1)(2x − 1) (e) (3x + 4)(3x − 4) (f) (5 − 3x)(5 + 3x)

(g) (x + 6)2 (h) (x + 11)2 (i) (8 + x)2

(j) (x − 3)2 (k) (x − 7)2 (l) (x − 12)2

(m) (2x + 5)2 (n) (3x + 2)2 (o) (7x − 2)2

4 Expand and simplify each of the following.

(a) x(x + 6)(x + 4) (b) x(x − 1)(x + 5) (c) x(x + 1)(x − 4)

(d) -x(x − 3)(x + 2) (e) -3x(x − 2)(x − 1) (f) -4x(x − 8)(x + 3)

(g) (x + 4)(x + 2)(x + 5) (h) (x + 2)(x − 5)(x − 1) (i) (x − 3)(x − 6)(x − 2)

(j) (x − 3)(x + 3)(x + 5) (k) (x + 7)(x − 2)(x + 2) (l) (2x + 1)(x + 4)(x − 4)

(m) (x − 3)(x + 4)(x + 4) (n) (x + 2)(x + 2)(x − 1) (o) (x − 6)(x + 5)(x − 6)

Understanding 5 The expression (x + 1)(x + 3)(x − 2) is not the same as:

A (x + 3)(x + 1)(x − 2) B (x − 3)(x + 1)(x + 2)

C (x + 1)(x − 2)(x + 3) D (x − 2)(x + 1)(x + 3)

6 Expand and simplify the following.

(a) (4x − 3y)(2x + 5y) (b) (5x + 2y)(7x + 6y) (c) (3x − 4y)(5x − 8y)

(d) (2x + 7y)(2x + 7y) (e) (7x + 6y)(7x + 6y) (f) (5x + 8y)(5x + 8y)

(g) (4x − y)(4x − y) (h) (6x − 5y)(6x − 5y) (i) (9x − 7y)(9x − 7y)

(j) (5y − x)(5y + x) (k) (3x − y)(3x + y) (l) (2x + 3y)(2x − 3y)

(m) x(2x − 1)(3x + 5) (n) x(3x + 2)(5x − 6) (o) 3x(4x + 7)(2x + 3)

(p) -x(3x − 4)(5x + 4) (q) -2x(5x + 3)(3x + 4) (r) -4x(2x − 3)(3x − 5)

7 Find the area, in terms of x, of each of these figures.

(a) (b) (c)

2

3

4

(x – 4) mm

(3x + 1) mm

(2x + 3) cm

(x – 5) m

(x + 5) m


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8 (a) Expand x(x + a)(x + b).

(b) Use your answer to (a) to:

(i) write down the coefficient of x2

(ii) write down the coefficient of x

(iii) show that x(x + a)(x + b) = x3 − 4x2 − 12x, when a = -6 and b = 2.

(c) For the expression x(x + 25)(x − 4):

(i) write down the values of a and b

(ii) use the expansion found in (a) to expand x(x + 25)(x − 4).

9 (a) If x is the second of three consecutive even numbers, write an expression for the product of the three numbers.

(b) Expand the expression in (a).

(c) Why is it simpler to make x the second and not the first number?

10 (a) Draw a rectangle of length (x + a) units and width (x + b) units.

(b) Write an expression for the area of the rectangle in (a) in factorised form.

(c) Expand and simplify your expression in (b).

(d) Draw a square of side length x units. Extend its length by a units and the width by b units to form four small rectangles.

(e) Add the areas of the four smaller rectangles and compare this answer to part (c).

11 (a) Write down the dimensions of the shaded rectangle.

(b) Write the binomial product used to find the area of the shaded region, then expand and simplify.

(c) Find the area of the square of side x cm and the rectangle of length 5 cm and width x cm.

(d) Find the unshaded area, writing it as the sum of the areas of the two rectangles.

(e) Use your answers in (c) and (d) to find the area of the shaded region in simplest form.

(f) Compare your answers in (b) and (e).

Reasoning 12 A square green tile of side length 36 cm has a small blue square of side length x cm painted

on it.

(a) Write the expression (in cm2) for the area of the tile that is not blue.

(b) Write this expression as a product of factors.

(c) Use this product to find the area that remains green if the blue square has a side of 6 cm.

13 (a) Expand and simplify (x − a)2 − a2.

(b) Write down the expansion for x(x − 2a).

(c) What can you conclude about (x − a)2 − a2 and x(x − 2a) from your answers in (a) and (b)?

(d) A rectangular lawn has a length of x m and a width of (x − 2a) m. A square lawn of length (x − a) m has a square-based fountain of side length a m in the centre. If the price of turfing the lawn is based only on its area, is one lawn more expensive to lay with turf than the other? Justify your answer.

(e) If the square lawn has side length 40 m, the fountain has a side length of 3 m and the price of turf is $12/m2, what is the total cost of laying the lawn?

x cm

x cm 5 cm

2 cm

x

x

36 cm

36 cm


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14 The Holmes family own a property at the beach. Their land is x m wide and extends down to the water’s edge. At low tide, the length of the land is 55 m, as shown in the diagram at right.

As the tide comes in, the length is reduced by y m.

(a) Write an expression for the length of the land at any time of day.

(b) Write an expression for the area of the land at any time of day.

(c) What is the area of the land at low tide?

(d) The family have decided to subdivide the land by reducing their land width by 30 m. Write an expression for the area of the remaining land at any time of the day.

(e) Expand the expression.

(f) If the tide comes in 10 m at high tide, what is the area of land at high tide?

(g) What is the difference between the area of land at low tide before the subdivision and at high tide after the subdivision?

15 Consider this box.

(a) Write an expression (in cm3) for the volume of the box in factorised form.

(b) Find the volume when x is 4 cm.

(c) Show that the volume of the box is 72 cm3, when x = 5.

(d) Write the expression (in cm3) for the volume in expanded form. Use this to calculate the volume when x = 5.

(e) State the dimensions (length, width and height) of the box when the volume is 72 cm3.

(f) What is the smallest possible integer value of x so that the box exists?

(g) Find the volume of the box when x is the value found in part (f).

16 Decide whether (x2 + a2)3 can equal zero for some value(s) of x and a other than x = 0, a = 0.

Open-ended 17 Find two different sets of values for a, b, c and d so that the expanded, simplified form of

(ax + b)(cx + d) is a binomial expression.

18 (a) Expand and simplify (x + y)2 + (x − y)2.

(b) Find two different sets of values for x and y so that the expression obtained in part (a) is the square of a whole number.

19 Find two different whole numbers representing x so that the value of the expression (4 − 2x)(x + 3)(5x − 10) is:

(a) positive (b) negative (c) zero.

x m

55 m

low tide

55 m

30 m

high tide

x m

y m

(x + 3) cm

(x – 2) cm

(x – 2) cm

A binomial expression contains two terms.

Hint

12--- 1

2---


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Factorising using common factorsFactorising is the opposite or reverse process to expanding. In the previous section we converted expressions from factorised form (with brackets) to expanded form (without brackets).

For example, 5(x + 3) = 5 × x + 5 × 3= 5x + 15

factorised form → expanded form

In other words, 5(x + 3) can be expanded to 5x + 15. In reverse, 5x + 15 can be factorised to 5(x + 3). This involves finding the highest common factor (HCF) of the terms and writing it outside a set of brackets that contains the remaining terms.

A common factor of -1 may also be taken out.

You can check your answer by expanding the brackets in your answer to get back to the original expression.

Worked Example 5Factorise:

(a) 15xy − 9y (b) 8x2 + 12x (c) -12a2b − 9ab2

Thinking(a) Find the HCF of the terms (3y). Write it in

front of the brackets. Divide each term by the HCF and write the quotient inside the brackets.

(a) 15xy − 9y= 3y(5x − 3)

(b) Find the HCF of the terms (4x). Write it in front of the brackets. Divide each term by the HCF and write the quotient inside the brackets.

(b) 8x2 + 12x= 4x(2x + 3)

(c) Find the HCF of the terms (3ab). Write it in front of the brackets. Take out the factor of -1 also. Divide each term by the negative of the HCF and write the quotient inside the brackets. Remember to change signs inside the brackets.

(c) -12a2b − 9ab2

= -3ab(4a + 3b)

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Sometimes the HCF is an expression of two or more terms.

Factorising four terms by grouping in pairsExpressions with four terms may factorise by grouping in pairs. Take out a common factor from each pair to give two terms. Then take out a common factor from each term.

We sometimes refer to this as a ‘two plus two’ factorisation.

Worked Example 6Factorise k(m + 3) + 5(m + 3).

ThinkingFind the HCF of the terms. (Here, it is (m + 3).) Write it in front of the brackets. Divide each term by the HCF and write this quotient inside the brackets.

k(m + 3) + 5(m + 3)= (m + 3)(k + 5)

Worked Example 7Factorise each of the following expressions by grouping in pairs.

(a) 2ax − 8ay − cx + 4cy (b) uw + vx + xu + vw

Thinking(a) 1 Factorise the first two terms by

finding the HCF. (Here, it is 2a.) Factorise the next two terms by finding the HCF. (Here, it is -c.)

Don’t forget to change signs when required.

(a) 2ax − 8ay − cx + 4cy= 2a(x − 4y) − c(x − 4y)

2 Find the new HCF and write it outside the brackets. (Here, it is (x − 4y).) Divide each term by the HCF and write the quotient inside the brackets.

= (x − 4y)(2a − c)

(b) 1 Rearrange the pairs so there is a common factor in each pair.

(b) uw + vx + xu + vw= uw + xu + vw + vx

2 Find and take out the HCF from the first two terms (here, it is u) and the HCF from the next two terms (here, it is v).

= u(w + x) + v(w + x)

3 Find the new HCF and write it outside the brackets. (Here, it is (w + x).) Divide each term by the HCF and write the quotient inside the brackets.

= (w + x)(u + v)

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Factorising using common factors

Fluency1 Factorise each expression.

(a) 4x + 8 (b) 18x − 15 (c) 24x − 9

(d) xy + 7x (e) 12kx + 21k (f) 32ab + 20b

(g) 7x2y + 7y2 (h) 9x2 + 3xy (i) 48m2 + 8m

(j) -6x − x2 (k) -4x + x2 (l) -24a2 + 10a

(m) 6ab − 18a2b (n) 40x2y + 16xy2 (o) 10cd2 − 2cd

2 Factorise each expression.

(a) x(k + 1) + 4(k + 1) (b) a(x − 4) + 6(x − 4)

(c) m(n − 7) + 2(n − 7) (d) x(p + 3) − 9(p + 3)

(e) n(q − 2) − 7(q − 2) (f) x(u − 4) − 5(u − 4)

(g) d(c + 5) + (c + 5) (h) a(d + 2) − (d + 2)

(i) k(m + 7) − (m + 7) (j) x(x − 3) + 5(x − 3)

3 Factorise each of the following expressions by grouping in pairs.

(a) xy + 3y + 5x + 15 (b) ab + 7b + 2a + 14 (c) mn + kn + 4m + 4k

(d) xy − y + 3x − 3 (e) 3a − 3b + ac − bc (f) km − 4m + 2k − 8

(g) 15ab − 10b − 12a + 8 (h) 6ab − 4b − 15a + 10 (i) 8mn − 6n − 12m + 9

(j) vx − vy − ax + ay (k) pq − rq − np + nr (l) gh − fh − dg + df

(m) m2n + np + m2t + pt (n) a2b + ab2 − ac − bc (o) d2k − dk − dr + r

(p) 27xy − 10 + 18y − 15x (q) 36a + 5c − 4ac − 45 (r) 25vw + 5v − 15w − 3

(s) 32a3c + 15b − 12abc − 40a2 (t) 21d2 + 20fh − 12dh − 35df (u) m2nr − 20p + 4mpr − 5mn

4 Answer TRUE or FALSE to each of these statements.

(a) 8a + 4ab = 4a(2 + b) (b) 2x − 6xy = 2x(1 − y)

(c) mn + 3n − 2m − 6 = (m − 3)(n + 2) (d) x2 − 4x + x − 4 = (x − 4)(x + 1)

(e) 3x2 + 15x − 2x − 10 = (x − 2)(3x + 5) (f) 4x2 + 20x − 3x − 15 = (x − 5)(4x + 3)

5 When factorised, 16f 2g − 12gf becomes:

A 2fg(8f − 6) B 4f 2g(4g − 3f ) C 16fg( f −12) D 4fg(4f − 3)

6 The total area of the two shapes shown is:

A (a + 4)(b − c) B (a + c)(b − 4) C (a + b)(c − 4) D (a − c)(b + 4)

NavigatorQ1 Column 1, Q2 Column 1, Q3 Column 1, Q4, Q6, Q7, Q8, Q9, Q11, Q13

Q1 Column 2, Q2 Column 2, Q3 Column 2, Q4, Q5, Q6, Q7, Q8, Q9, Q11, Q12, Q13

Q1 Column 3, Q2 Column 2, Q3 Column 3, Q4, Q5, Q6, Q7, Q8, Q9, Q10, Q11, Q12, Q13

2.2

Answerspage 720

5

6

7

b – 4

a

b – 4

c


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2.2


7 10dk + 25d − 2k − 5 factorises to:

A (2k + 5)(5d − 1) B 5d(2k + 5) C (2k − 5)(5d + 1) D (10k + 1)(d − 5)

Understanding8 Mandy has 24y cents. She buys x kg of potatoes that cost 36 cents per kilogram.

Express (in cents) in factorised form how much money Mandy has now.

9 Factorise the expression for the square of a given number subtracted from eight times that number.

10 Five small bulldog clips are added to a box containing p small bulldog clips. Five large bulldog clips are added to another box that contains p large bulldog clips. Each small clip costs a cents and each large clip costs b cents.

(a) Give the expression (in cents) for the combined cost of the clips.

(b) Factorise the expression in part (a).

(c) If the cost of one large clip is twice the cost of one small clip write a factorised expression for the total cost of the clips in terms of a and p.

11 (a) Draw and label a rectangle that has a length of p m and a width of r m.

(b) The length is reduced by a m. Show this on your diagram and find the new area as the difference of two areas.

(c) The width of this shortened rectangle is now increased by b m. Show this on your diagram. Label the dimensions and find an expression for the added area.

(d) Show that the area of the newly formed rectangle is (pr − ar + pb − ab) m2.

(e) Find the area as a product of the new length and width.

(f) Comment on your answers to parts (d) and (e).

Reasoning12 The volume of a sphere of radius r is and the

volume of a cone of base radius r and height h is

Renata is holding an ice-cream cone, with a scoop of ice-cream on top. The ice-cream can be considered to be hemispherical (half a sphere).

(a) Give an expression for the total volume of ice-cream (m3), including the ice-cream in the cone and the scoop on top, if the ice-cream filled the cone.


(c) If r = 3 cm, h = 10 cm and π = 3.142, what is the total volume? (Give your answer correct to two decimal places.)

(d) Assume that there is no ice-cream inside the cone, just the scoop on top, and Renata lets it melt inside the cone. If the height of the cone is 2r, use your expression in part (b) to show that the total volume is twice the volume of the cone and therefore all the melted ice-cream will fill the cone exactly.

Open-ended13 A paver has the shape and dimensions shown.

(a) Write an expression (in cm2) for the total surface area.


(c) Find two sets of values for x and y so that the volume (in cm3) has numerically the same value as the surface area. The values do not have to be integers.

h cm

2r cm

4πr3

3-----------

πr2h3

----------- .

x cm

y cm

x cm


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2 Algebra 65

Factorising monic quadratic expressionsA quadratic expression has the form of ax2 + bx + c.

In a quadratic expression the highest power of the variable is 2. In a monic quadratic expression, the coefficient of the squared variable is 1. For example, x2 + 5x − 4, 6 − 3a2, h2 and 4n2 + 7n are all quadratic expressions. Of these, only x2 + 5x − 4 and h2 are monic quadratic expressions.

Factorising quadratic expressions with four termsIn the previous section we factorised four terms by grouping in pairs into binomial products. If the first term is a squared variable and the two middle terms contain the same variable and the last term is a constant we may be able to use grouping in pairs to factorise these expressions into binomial products.

Factorising quadratic trinomialsA quadratic trinomial consists of three terms. For example, x2 + 5x − 4 is a quadratic trinomial. The monic quadratic trinomials that we will factorise in this section will all be in the form of ax2 + bx + c where a = 1. Not all quadratic expressions in this form will factorise into binomial products.

Consider x2 + 3x − 5x + 5:

We can take out a common factor from each pair to give x(x + 3) − 5(x − 1) but it will not factorise any further. Why do some expressions with four terms factorise fully into binomial products whereas others do not?

Worked Example 8Factorise x2 − 4x + 5x − 20.

Thinking1 Write down the expression. Check if

there is a factor common to all terms. (None.)

x2 − 4x + 5x − 20

2 Group ‘in pairs’ by looking for a common factor in the first and the second pair of terms (x2 − 4x has a common factor of x) (5x − 20 has a common factor of 5). Take out the factor from each pair.

= x(x − 4) + 5(x − 4)

3 Take out the factor common to both groups (x − 4).

= (x − 4)(x + 5)

8


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The firefighter’s ladder

Equipment required: 1 brain, calculator

The Big QuestionCan we predict how the height of the top of a ladder will change for a small change in the distance of the foot of the ladder from the wall?

EngageA firefighter has to get into a building through a window by leaning his ladder against the wall. Given the length of his ladder and the height of the window, he will need to calculate how far from the wall he must put the bottom of his ladder.

Give exact answers for each of the following questions.

1 (a) For a 10 m ladder, calculate each of the following heights (h).

(i) (ii)

(iii) (iv)

(b) If the ladder is 10 m long, find the value of h, when the distance from the base of the wall to the foot of the ladder is equal to the height to the top of the ladder on the wall.

ExploreGive exact answers for each of the following questions.

2 (a) If the distance of the foot of the ladder to the wall and the height of the top of the ladder on the wall are both equal to 7 m, (h = 7 m), how long would the ladder have to be?

(b) Find the difference between the length of this ladder and 7 m.

h10 m

5 m

h10 m

6 m

h10 m

7 m

h10 m

8 m

h10 m

h

Investigation


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2 Algebra 97

(c) How far could this ladder move in towards the wall before it was completely upright?

(d) How far could the bottom of the ladder slip away from its position 7 m from the base of the wall before it was flat on the ground?

(e) If the base of the ladder slips p m away from the base of the wall and the top of the ladder travels a distance of n m down the wall, what is the largest possible values for:

(i) p (ii) n

(f) If the base of the ladder slips p m away from the base of the wall and the top of the ladder travels a distance of n m down the wall, show that (p + 7)2 + (7 − n)2 = 98.

(g) (i) Rearrange this formula and simplify to show that (7 − n)2 = 49 − 14p − p2.

(ii) Hence, show that n = 7 −

(iii) Why do we reject n = 7 +

(h) Evaluate n for all possible integer values of p.

(i) Evaluate n for p = -16.

ExplainUse exact values in your explanations for each of the following questions.

3 (a) Does the ladder slip down the wall the same distance it moves away from the base of the wall? Explain why or why not. If not, does it slip further down or not as far down the wall than the distance away from it?

(b) Explain what happens to the distance the ladder will move up the wall when the ladder is moved closer to the wall.

(c) In 2(h), we could only evaluate n for integer values of p ≤ 2. Explain why.

(d) Why can we evaluate n for p = -16 m when this value of p is not possible? What does p = -16 m mean?

Elaborate4 (a) If h, the distance from the base of the wall to the

bottom of the ladder, is equal to the height to the top of the ladder on the wall, find an equation for n, the distance the top of the ladder moves down the wall when the foot of the ladder moves a distance of p away from the base of the wall.

(b) Use an equation to show that p can only be equal to n for p = n = 0.

(c) Describe what happens to the distance the ladder can reach up a wall as the bottom of the ladder slips away from the base of the wall and as the bottom of the ladder is pushed closer to the base of the wall.

Evaluate5 (a) Did you find visualising the ladder moving

up and down the wall a challenge? Did the diagrams help you?

(b) Did you find working with unknown variables more difficult than working with actual values? Did working with actual values help you to understand the situation? Explain why or why not.

(c) Why were you asked to give exact answers? Would it have been easier to understand if you had evaluated lengths and distances correct to two decimal places? Explain.

Extend6 If the distance from the foot of the ladder to the base

of the wall is a − b and the distance up the wall to the top of the ladder is a + b, show that the length of the ladder can be written as

Refer to your answer in part (e)(ii).

Hint

49 14p– p2– .

49 14p– p2– ?

• Make a table.

• Look for a pattern.

Strategy options

Did you find n was the same for one of the values of the p-values you used in Question 2(h)? If so, which value?

Hint

The length of the ladder does not change.

Hint

2 a2 b2+( ).


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Chapter review D.I.Y. Summary

Copy and complete the following using the words and phrases from the list where appropriate to write a summary for this chapter. A word or phrase may be used more than once.

1 ax2 + bx + c is called a . When the coefficient of x2 is 1 the expression is a .

2 The first thing to do to factorise 50x2 − 2y2 is to take out the . We can then use the method to factorise the expression.

3 To factorise x2 − 4x − 6 as (x − 2 − (x − 2 + we use a method called .

4 We to write the expression 2ax + 2ay + 3x + 3y as a product of factors (2a + 3)(x + y).

5 (3t − 2)2 is an example of a .

6 (3a + b)(4 − 2a) is called a .

7 x2 − 9 is an example of the .

Fluency1 Expand and simplify the following.

(a) 5(2x + 7) (b) x(x − 3) + 2(x − 3)

(c) (x + 6)(x + 4) (d) (4x − 1)(2x + 3)


(a) (x + 6)2 (b) (2x − 3)2

(c) (5 − x)(5 + x) (d) (3x + 7)(3x − 7)


(a) 4(x − 7)(x + 2) (b) -3x(x + 8)(x − 3)

(c) (x − 1)(x + 5)(x − 6) (d) (2x + 3)(2 − x)(x + 4)

4 Factorise these expressions.

(a) 7x + 21 (b) 12ab − 8a (c) 18x2 + 24x


(a) d(c − 4) + 6(c − 4) (b) x(x + 2) − 3(x + 2)

(c) 5a(a + 1) − 2(a + 1) (d) 2y(3y − 5) − 7(3y − 5)


(a) x2 + 11x + 30 (b) x2 − 10x − 24

algebraic fractions factorise by grouping in pairs Perfect Square

binomial product factorising quadratic expression

completing the square highest common factor (HCF) quadratic trinomial

Difference of Two Squares monic quadratic

2Key Words

10) 10)

2.1

2.1

2.1

2.2

2.2

2.3


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1012 Algebra

7 Factorise each of these quadratic trinomials.

(a) x2 + 10x + 25 (b) x2 − 18x + 81

(c) 9x2 − 24x + 16 (d) 4x2 + 4x + 1


(a) x2 − 9 (b) 25 − y2 (c) 4x2 − y2

(d) 49x2 − 16 (e) 36x2 − 81y2 (f) 64 − 9x2

(g) (2x − 5)2 − 4 (h) (x − 5)2 − (x + 8)2 (i) (x + 2)2 − (x − 3)2

(j) x2 − 3 (k) (x + 7)2 − 5 (l) (x − 4)2 − 7


(a) x2 + 6x − 2 (b) x2 − 10x + 23

(c) x2 − 4x − 7 (d) x2 + 8x + 13

10 Simplify the following.

(a) (b) (c) (d)


(a) (b)

(c) (d)


(a) (b) (c)

(d) (e) (f)

13 Rearrange each of the following to make the pronumeral in brackets the subject.

(a) A = lb [b] (b) A = [l]

(c) SA = 4πr2 [r] (d) a = x2y − z [x]

Understanding14 If 9x2 − 12x + 4 is written as (ax + b)2, state the values of a and b.

15 In January, John’s weight was x kg. In April his weight was (x + a) kg and in September it was (x + b) kg.

(a) If the product of John’s weight in April and September is x2 + x − 30, determine the values of a and b.

(b) By how many kilograms had John's weight increased by April?

(c) What was John’s overall weight gain/loss by the end of September?

(d) What happened to his weight from April to September?

16 The heights in centimetres of four people are ax, 2x, ay and 2y.

(a) Write down the expression for their total height.


(c) If x = 65, y = 75 and a = 3, use your factorised expression to find the average height of the four people.

2.4

2.4

2.5

2.618x45-------- 2 a 3–( )

6-------------------- 10x x 1–( )

8x------------------------- 12ab

4b------------

2.6x 4+( ) x 3–( )x 3–( ) x 2+( )

--------------------------------- 2x 1–( ) x 1+( )x 7+( ) x 7–( )

------------------------------------ 3x 7+( ) x 7+( )x 1–( ) 2x 1–( )

-------------------------------------×

4x2 4x– 1+2x 1–

----------------------------- 2x2 8x–x2 3x+

-------------------- x2 16–x2 8x 15+ +-----------------------------÷

2.62x3----- x

2---+ x 1+

5----------- x 2–

3-----------+ 3x 2–

10-------------- x 5+

6-----------–

6x7----- 3y

2------+ x 5+

4----------- y 2+

3------------+ x 6–

5----------- y 3–

10-----------–

2.72π l

g--


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17 (a) Which of the following methods of factorising is the most appropriate for each of the expressions (i) to (vi) below?

A factorising using common factors

B factorising using the Difference of Two Squares rule

C factorising by grouping in pairs

D factorise by grouping (3 and 1)

E factorising quadratic trinomials (multiply to give … , add to give …)

F factorising by completing the square

(i) x2 + 5x − 24 (ii) 25a2 − b2

(iii) x2 + 6x + 9 − y2 (iv) x2 + 12x + 34

(v) 16x2 − 25x (vi) 4x − 4y + x2 − y2

(b) Factorise each expression in part (a).

18 A house has had a rectangular extension added. The extension has a length of (x + 6) m and a width of (x + 4) m. An area 2 m wide and 4 m long is to be an untiled verandah. The remainder is to be a tiled family room. The cost of tiling a family room is $18 per square metre.

(a) Write an expression for the area of the family room in factorised form.

(b) Write an expression for the cost of tiling the family room.

(c) Find the cost of tiling the family room if x is 2 m.

19 Find, in expanded form, the square of the sum of 4x and 3y.

Reasoning20 (a) Write an expression (in m2) for the area of

the kitchen bench surface shown at right. Simplify the expression as far as possible.

(b) Factorise the expression for the area.

(c) Find the area of the kitchen bench surface when the width of the bench, x, is:

(i) 1 m

(ii) 0.5 m

(iii) 0.75 m

21 A biscuit box has a length 14 cm greater than its height and a width 9 cm greater than its height.

(a) Draw a diagram showing this information. Use h to represent height.

(b) Write an expression (in factorised form) to represent the volume of the biscuit box.

(c) Expand and simplify the expression for volume.

(d) Use your answer for part (b) to find the volume of the biscuit box if its height is 6 cm.

(e) Use your answer for part (c) to find the volume of the biscuit box if its height is 6 cm.

(f) Which form of the volume expression (factorised or expanded) did you find easier to work with? Explain your answer.

(x + 4) m

4 m

2 m

(x + 6) m

Verandah

Familyroom

x m

3 m

2 m

x m


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1032 Algebra

22 A circular disc of metal expands when it is heated. When it is heated to a particular temperature, its radius increases by 2 mm. When cooled to a particular temperature below room temperature, the disc contracts and its radius decreases by 2 mm. (Do not substitute for π in any parts of this question.)

(a) If the disc has a radius of x mm at room temperature (20°C), write down its area at room temperature.

(b) In terms of x, what is the radius of the heated disc?

(c) Hence, write an expression for the area of this larger disc.

(d) In terms of x, what is the radius of the cooled disc?

(e) Hence, write an expression for the area of this smaller disc.

(f) Write an expression for the difference between the area of the larger disc and that of the smaller disc.

(g) Is there a common factor that can be taken out of this expression for the difference in area? What is it?

(h) Factorise this expression for the difference in area.

(i) If the original disc (at 20°C) had a radius of 10 mm, find the difference in area between the larger and smaller discs.

(j) If the difference in area between the larger and smaller discs is 136π mm2, find the radius of the disc at 20°C.

23 An exercise book is (8x + 3) cm long and (2y + 7) cm wide.

(a) Write an expression (in cm2) for the area of the book.

(b) What is the expression for the area if y = 4x − 5?

(c) Expand and simplify the expression in part (b).

(d) (i) In terms of x, what value of y will result in the expression for the area being a perfect square?

(ii) Expand and simplify the perfect square in (i).

24 The number of private cars observed during a period of x days is modelled by the expression x2 + p2x where p is the number of passengers. The number of taxis observed with p passengers during y days is given by p2y − y2.

(a) Write the expression for the total number of cars and taxis.


(c) Explain why x must be greater than y if the number of passengers is zero.

(d) (i) How many vehicles were there in total if x = y?

(ii) It is found that 2px of these vehicles were white. Write, in factorised form, the number of non-white vehicles.

Disc expands when heated Disc contracts when cooled

(2y

+ 7)

cm

(8x + 3) cm


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