Twitter: @Owen134866 · (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 ≠

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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







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







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…


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!


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


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)


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)


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…


x2 + 10x + 16


+1 +16

-1 -16

+2 +8

-2 -8

+4 +4

-4 -4

Which pair adds to give +10?

(x + 2)(x + 8)



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)



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




ax2 + bx + c



1E

Examples

b)2 4 5x x



Add to give ‘b’

( 5)( 1)x x




ax2 + bx + c



1E

Examples

c)2 25x



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)




ax2 + bx + c



1E

Examples

d)2 24 9x y



Add to give ‘b’

(2 3 )(2 3 )x y x y




ax2 + bx + c



1E

Examples

d)25 45x



Add to give ‘b’

Sometimes, you need to remove a ‘common factor’ first…

25( 9)x

5( 3)( 3)x x


• 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


2x2 - 5x - 3


-3 +1

+3 -1

One of the values to the left will be doubled when the brackets are expanded

(2x + 1)(x - 3)


-6 +1

-3 +2

+6 -1

+3 -2 The -3 doubles so it must be on the opposite side to the ‘2x’


2x2 + 13x + 11


+11 +1

-11 -1

One of the values to the left will be doubled when the brackets are expanded

(2x + 11)(x + 1)


+22 +1

+11 +2

-22 -1

-11 -2 The +1 doubles so it must be on the opposite side to the ‘2x’


3x2 - 11x - 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)


+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



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





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





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





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





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


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



number.


rules:


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



number.


rules:


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


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


𝑎+ 𝑏, multiply the

numerator and denominator by 𝑎 − 𝑏


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





𝑎, multiply the








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


Simplify





𝑎, multiply the








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


Simplify

Divide all by 2

Twitter: @Owen134866 · (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 ≠

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