twitter: @owen134866 · (2 3 )(2 3 )x y x y algebraic expressions. factorising quadratics a...
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Twitter: @Owen134866
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Prior Knowledge Check
1) Simplify:
a) 4𝑚2𝑛 + 5𝑚𝑛2 − 2𝑚2𝑛 + 𝑚𝑛2 − 3𝑚𝑛2
b) 3𝑥2 − 5𝑥 + 2 + 3𝑥2 − 7𝑥 − 12
2) Write as a single power of 2
a) 25 × 23 b) 26 ÷ 22
c) 23 2
3) Expand:
a) 3(𝑥 + 4) b) 5(2 − 3𝑥)
c) 6(2𝑥 − 5𝑦)
4) Write down the highest common factor of:
a) 24 and 16 b) 6𝑥 and 8𝑥2
c) 4𝑥𝑦2 and 3𝑥𝑦
5) Simplify:
a) 10𝑥
5b)
20𝑥
2
c) 40𝑥
24
2𝑚2𝑛 + 3𝑚𝑛2
6𝑥2 − 12𝑥 − 10
28
24
26
3𝑥 + 12
10 − 15𝑥12𝑥 − 30𝑦
82𝑥
𝑥𝑦
2𝑥 10𝑥
5𝑥
3
Algebraic Expressions
You can use the laws of indices to simplify powers of the same base
𝑎𝑚 × 𝑎𝑛 = 𝑎𝑚+𝑛
𝑎𝑚 ÷ 𝑎𝑛 = 𝑎𝑚−𝑛
𝑎𝑚 𝑛 = 𝑎𝑚𝑛
𝑎𝑏 𝑛 = 𝑎𝑛𝑏𝑛
1A
a) 𝑥2 × 𝑥5 = 𝑥7
b) 2𝑟2 × 3𝑟3 = 6𝑟5
c) 𝑏7
𝑏4= 𝑏3
d) 6𝑥5 ÷ 3𝑥3 = 2𝑥2
e) 𝑎3 2 × 2𝑎2
= 𝑎6 × 2𝑎2
= 2𝑎8
f) 3𝑥2 3 ÷ 𝑥4
= 27𝑥6 ÷ 𝑥4
= 27𝑥2
Algebraic Expressions
You can use the laws of indices to simplify powers of the same base
𝑎𝑚 × 𝑎𝑛 = 𝑎𝑚+𝑛
𝑎𝑚 ÷ 𝑎𝑛 = 𝑎𝑚−𝑛
𝑎𝑚 𝑛 = 𝑎𝑚𝑛
𝑎𝑏 𝑛 = 𝑎𝑛𝑏𝑛
1A
Expand and simplify if possible
= −21𝑥2 + 12𝑥
a) −3𝑥(7𝑥 − 4)
= 3𝑦2 − 2𝑦5
b) 𝑦2(3 − 2𝑦3)
= 12𝑥2 − 8𝑥3 + 20𝑥4
c) 4𝑥(3𝑥 − 2𝑥2 + 5𝑥3)
= 10𝑥2 + 6𝑥 − 10𝑥 − 15
d) 2𝑥 5𝑥 + 3 − 5(2𝑥 + 3)
= 10𝑥2 − 4𝑥 − 15
Algebraic Expressions
You can use the laws of indices to simplify powers of the same base
𝑎𝑚 × 𝑎𝑛 = 𝑎𝑚+𝑛
𝑎𝑚 ÷ 𝑎𝑛 = 𝑎𝑚−𝑛
𝑎𝑚 𝑛 = 𝑎𝑚𝑛
𝑎𝑏 𝑛 = 𝑎𝑛𝑏𝑛
1A
Simplify
= 𝑥4 + 𝑥
a) 𝑥7+𝑥4
𝑥3
=3𝑥
2
b) 3𝑥2−6𝑥5
2𝑥
=𝑥7
𝑥3+𝑥4
𝑥3
=3𝑥2
2𝑥−6𝑥5
2𝑥
− 3𝑥4
= 4𝑥5
c) 20𝑥7+15𝑥3
5𝑥2=20𝑥7
5𝑥2+15𝑥3
5𝑥2
+ 3𝑥
If you have a single term as the
denominator, you can simplify the numerator
terms separately…
Algebraic Expressions
To find the product of two expressions you multiply each term in one expression by each term in
the other expression
1B
(x + 4)(x + 7)
x2 + 4x + 7x + 28
x2 + 11x + 28
(2x + 3)(x – 8)
2x2 + 3x – 16x – 24
2x2 – 13x - 24
+ 28+ 7x+ 7
+ 4xx2x
+ 4x
- 24- 16x- 8
+ 3x2x2x
+ 32x
There are various methods for doing this,
all are ok!
Algebraic Expressions
To find the product of two expressions you multiply each term in one expression by each term in
the other expression
If you have more than two brackets, just multiply any 2 first, and then
multiply the answer by the next one
1B
Expand 𝑥 + 4 2𝑥 − 1 𝑥 + 3
𝑥 + 4 2𝑥 − 1 𝑥 + 3
= (2𝑥2 + 7𝑥 − 4)(𝑥 + 3)
= 2𝑥3 + 6𝑥2+ 7𝑥2 + 21𝑥− 4𝑥 − 12
= 2𝑥3 + 13𝑥2 + 17𝑥 − 12
Multiply the first pair of brackets
Multiply this new pair
Simplify
Algebraic Expressions
You can write expressions as products of their factors. This is
known as factorising.
If the terms have a common factor (or several), then the expression can be factorized into a single bracket
1C
3 9xa) 3( 3)x
Common Factor
3
2 5x xb) ( 5)x x x
28 20x xc) 4 (2 5)x x 4x
2 29 15x y xyd) 3 (3 5 )xy x y 3xy
23 9x xye) 3 ( 3 )x x y 3x
x2 + 3x 2+
You get the last number in a Quadratic Equation by multiplying the 2 numbers in the brackets
You get the middle number by adding the 2 numbers in the brackets
(x + 2)(x + 1)
Algebraic Expressions
x2 - 2x 15-
You get the last number in a Quadratic Equation by multiplying the 2 numbers in the brackets
You get the middle number by adding the 2 numbers in the brackets
(x - 5)(x + 3)
Algebraic Expressions
x2 - 7x + 12
Numbers that multiply to give + 12
+3 +4
-3 -4
+12 +1
-12 -1
+6 +2
-6 -2
Which pair adds to give -7?
(x - 3)(x - 4)
So the brackets were originally…
Algebraic Expressions
x2 + 10x + 16
Numbers that multiply to give + 16
+1 +16
-1 -16
+2 +8
-2 -8
+4 +4
-4 -4
Which pair adds to give +10?
(x + 2)(x + 8)
So the brackets were originally…
Algebraic Expressions
x2 - x - 20
Numbers that multiply to give - 20
+1 -20
-1 +20
+2 -10
-2 +10
+4 -5
-4 +5
Which pair adds to give - 1?
(x + 4)(x - 5)
So the brackets were originally…
Algebraic Expressions
Factorising Quadratics
A Quadratic Equation has the form;
ax2 + bx + c
Where a, b and c are constants and a ≠ 0.
You can also Factorise these equations.
REMEMBER
An equation with an ‘x2’ in does notnecessarily go into 2 brackets. You use 2 brackets when there are NO ‘Common Factors’
1E
Examples
a)2 6 8x x
The 2 numbers in brackets must:
Multiply to give ‘c’
Add to give ‘b’
( 2)( 4)x x
Algebraic Expressions
Factorising Quadratics
A Quadratic Equation has the form;
ax2 + bx + c
Where a, b and c are constants and a ≠ 0.
You can also Factorise these equations.
1E
Examples
b)2 4 5x x
The 2 numbers in brackets must:
Multiply to give ‘c’
Add to give ‘b’
( 5)( 1)x x
Algebraic Expressions
Factorising Quadratics
A Quadratic Equation has the form;
ax2 + bx + c
Where a, b and c are constants and a ≠ 0.
You can also Factorise these equations.
1E
Examples
c)2 25x
The 2 numbers in brackets must:
Multiply to give ‘c’
Add to give ‘b’
( 5)( 5)x x
(In this case, b = 0)
This is known as ‘the difference of two squares’
x2 – y2 = (x + y)(x – y)
Algebraic Expressions
Factorising Quadratics
A Quadratic Equation has the form;
ax2 + bx + c
Where a, b and c are constants and a ≠ 0.
You can also Factorise these equations.
1E
Examples
d)2 24 9x y
The 2 numbers in brackets must:
Multiply to give ‘c’
Add to give ‘b’
(2 3 )(2 3 )x y x y
Algebraic Expressions
Factorising Quadratics
A Quadratic Equation has the form;
ax2 + bx + c
Where a, b and c are constants and a ≠ 0.
You can also Factorise these equations.
1E
Examples
d)25 45x
The 2 numbers in brackets must:
Multiply to give ‘c’
Add to give ‘b’
Sometimes, you need to remove a ‘common factor’ first…
25( 9)x
5( 3)( 3)x x
Algebraic Expressions
• Expand the following pairs of brackets
(x + 3)(x + 4)
x2 + 3x + 4x + 12
x2 + 7x + 12
(2x + 3)(x + 4)
2x2 + 3x + 8x + 12
2x2 + 11x + 12
+ 12+ 4x+ 4
+ 3xx2x
+ 3x
+ 12+ 8x+ 4
+ 3x2x2x
+ 32x
When an x term has a ‘2’ coefficient, the rules are
different…
2 of the terms are doubled
So, the numbers in the brackets add to
give the x term, WHEN ONE HAS BEEN DOUBLED FIRST
Algebraic Expressions
2x2 - 5x - 3
Numbers that multiply to give - 3
-3 +1
+3 -1
One of the values to the left will be doubled when the brackets are expanded
(2x + 1)(x - 3)
So the brackets were originally…
-6 +1
-3 +2
+6 -1
+3 -2 The -3 doubles so it must be on the opposite side to the ‘2x’
Algebraic Expressions
2x2 + 13x + 11
Numbers that multiply to give + 11
+11 +1
-11 -1
One of the values to the left will be doubled when the brackets are expanded
(2x + 11)(x + 1)
So the brackets were originally…
+22 +1
+11 +2
-22 -1
-11 -2 The +1 doubles so it must be on the opposite side to the ‘2x’
Algebraic Expressions
3x2 - 11x - 4
Numbers that multiply to give - 4
+2 -2
-4 +1
+4 -1
One of the values to the left will be tripled when the brackets are expanded
(3x + 1)(x - 4)
So the brackets were originally…
+6 -2
+2 -6
-12 +1
-4 +3
The -4 triples so it must be on the opposite side to the ‘3x’
+12 -1
+4 -3
Algebraic Expressions
Algebraic Expressions
Indices can be negative numbers or fractions
𝑎1𝑚 = 𝑚 𝑎
𝑎𝑛𝑚 = 𝑚 𝑎 𝑛
𝑎−𝑚 =1
𝑎𝑚
𝑎0 = 1
1D
Simplify
a) 𝑥3
𝑥−3= 𝑥6
b) 𝑥1
2 × 𝑥3
2 = 𝑥2
c) 𝑥32
3 = 𝑥2
d) 3125𝑥6 = 125𝑥6
13
= (125)13 𝑥6
13
= 5𝑥2
Algebraic Expressions
Indices can be negative numbers or fractions
𝑎1𝑚 = 𝑚 𝑎
𝑎𝑛𝑚 = 𝑚 𝑎 𝑛
𝑎−𝑚 =1
𝑎𝑚
𝑎0 = 1
1D
Simplify
e) 2𝑥2−𝑥
𝑥5=2𝑥2
𝑥5−𝑥
𝑥5
=2
𝑥3−1
𝑥4
= 2𝑥−3 − 𝑥−4
Either of these forms is correct – check if the question asks for a specific one!
Simplify separately
Rewrite
Algebraic Expressions
Indices can be negative numbers or fractions
𝑎1𝑚 = 𝑚 𝑎
𝑎𝑛𝑚 = 𝑚 𝑎 𝑛
𝑎−𝑚 =1
𝑎𝑚
𝑎0 = 1
1D
Evaluate (work out the value of)
a) 91
2 = 9
= 3
b) 641
3 =364
= 4
c) 493
2 = 493
= 343
d) 25−3
2 =1
2532
=1
253 =
1
125
You can use a calculator for these, but you still need to be able to show the process, especially for algebraic versions
Algebraic Expressions
Indices can be negative numbers or fractions
𝑎1𝑚 = 𝑚 𝑎
𝑎𝑛𝑚 = 𝑚 𝑎 𝑛
𝑎−𝑚 =1
𝑎𝑚
𝑎0 = 1
1D
Given that 𝑦 =1
16𝑥2, express 𝑦
1
2 in the
form 𝑘𝑥𝑛 where 𝑘 and 𝑛 are constants
𝑦 =1
16𝑥2
𝑦12 =
1
16𝑥2
12
𝑦12 =
1
16
12
𝑥212
𝑦12 =
1
4𝑥
Rewrite based on the question
Each part is raised to a power ½
Simplify
Algebraic Expressions
Indices can be negative numbers or fractions
𝑎1𝑚 = 𝑚 𝑎
𝑎𝑛𝑚 = 𝑚 𝑎 𝑛
𝑎−𝑚 =1
𝑎𝑚
𝑎0 = 1
1D
Given that 𝑦 =1
16𝑥2, express 4𝑦−1 in the
form 𝑘𝑥𝑛 where 𝑘 and 𝑛 are constants
𝑦 =1
16𝑥2
4𝑦−1 = 41
16𝑥2
−1
4𝑦−1 = 41
16
−1
𝑥2 −1
4𝑦−1 = 4(16)(𝑥−2)
Rewrite based on the question
Each part is raised to a power -1, and will then
be multiplied by 4
Simplify
4𝑦−1 = 64𝑥−2
Simplify more
Algebraic Expressions
In 𝒏 is an integer that is not a square number, then 𝒏 is a surd. It is an example of an irrational
number.
Surds can be used to leave answers exact without rounding errors, and can be manipulated by using the following
rules:
𝑎𝑏 = 𝑎 × 𝑏
𝑎
𝑏=
𝑎
𝑏
1E
Simplify
a) 18 = 9 × 2
= 3 2
Make sure that what you write is clear…
3 2 and 32 are different!
b) 20
2=
4 × 5
2
=2 5
2
= 5
Find a factor which is a square number, which you
can then square root
Simplify the numerator
Simplify the whole fraction
Algebraic Expressions
In 𝒏 is an integer that is not a square number, then 𝒏 is a surd. It is an example of an irrational
number.
Surds can be used to leave answers exact without rounding errors, and can be manipulated by using the following
rules:
𝑎𝑏 = 𝑎 × 𝑏
𝑎
𝑏=
𝑎
𝑏
1E
Simplify
c) 5 6 − 2 24 + 294Try to find a
common factor= 5 6 − 2 4 6 + 49 6
= 5 6 − 4 6 + 7 6
= 8 6
Square roots can be worked out
Simplify
Algebraic Expressions
In 𝒏 is an integer that is not a square number, then 𝒏 is a surd. It is an example of an irrational
number.
Surds can be used to leave answers exact without rounding errors, and can be manipulated by using the following
rules:
𝑎𝑏 = 𝑎 × 𝑏
𝑎
𝑏=
𝑎
𝑏
1E
Expand and simplify if possible
a) 2 5 − 3Multiply out
= 5 2 − 6
b) 2 − 3 5 + 3Multiply out
= 10 + 2 3 − 5 3 − 9
= 10 − 3 3 − 3
= 7 − 3 3
Group together like terms. Calculate root 9
Simplify
Algebraic Expressions
If a fraction has a surd in the denominator, then it can be useful to rearrange it so that the denominator is a rational number. This is called
rationalising the denominator.
For fractions of the form 1
𝑎, multiply the
numerator and denominator by 𝑎
For fractions of the form 1
𝑎+ 𝑏, multiply the
numerator and denominator by 𝑎 − 𝑏
For fractions of the form 1
𝑎− 𝑏, multiply the
numerator and denominator by 𝑎 + 𝑏
1F
Rationalise
a) 1
3
1
3×
3
3
=3
3
b) 1
3+ 2
1
3 + 2×3 − 2
3 − 2
=3 − 2
3 + 2 3 − 2
=3 − 2
9 + 3 2 − 3 2 − 2
=3 − 2
7
Multiply so that the surd is removed from
the denominator
Multiply both numerator and denominator
Multiply out the brackets
Simplify
Algebraic Expressions
If a fraction has a surd in the denominator, then it can be useful to rearrange it so that the denominator is a rational number. This is called
rationalising the denominator.
For fractions of the form 1
𝑎, multiply the
numerator and denominator by 𝑎
For fractions of the form 1
𝑎+ 𝑏, multiply the
numerator and denominator by 𝑎 − 𝑏
For fractions of the form 1
𝑎− 𝑏, multiply the
numerator and denominator by 𝑎 + 𝑏
1F
Rationalise
c) 5+ 2
5− 2
5 + 2
5 − 2×
5 + 2
5 + 2
=5 + 2 5 + 2
5 − 2 5 + 2
=5 + 10 + 10 + 2
5 + 10 − 10 − 2
=7 + 2 10
3
Multiply both numerator and denominator
Multiply out the brackets
Simplify
Algebraic Expressions
If a fraction has a surd in the denominator, then it can be useful to rearrange it so that the denominator is a rational number. This is called
rationalising the denominator.
For fractions of the form 1
𝑎, multiply the
numerator and denominator by 𝑎
For fractions of the form 1
𝑎+ 𝑏, multiply the
numerator and denominator by 𝑎 − 𝑏
For fractions of the form 1
𝑎− 𝑏, multiply the
numerator and denominator by 𝑎 + 𝑏
1F
Rationalise
d) 1
1− 32
Multiply out the brackets first
=1
1 − 3 1 − 3
=1
4 − 2 3
=1
4 − 2 3×4 + 2 3
4 + 2 3
=4 + 2 3
(4 − 2 3) 4 + 2 3
=4 + 2 3
16 + 8 3 − 8 3 − 4 9
=4 + 2 3
4
=2 + 3
2
Multiply to cancel the surds
Multiply out the brackets
Simplify
Divide all by 2