FACTORISING
Higher Tier
COMMON FACTORS

Factorising is the reverse of expanding brackets.

When an expression is factorised, brackets are put back in.

An expression is factorised by taking out common factors.
Examples
1.
3a + 6 = 3(a + 2)
[3 is the common factor of each term]
2.
xy + x = x(y + 1)
[x is the common factor of each term]
3.
5ab + 10b = 5b(a + 2)
[5b is the common factor of each term]
4.
2x² + 12x = 2x(x + 6)
[2x is the common factor of each term]
5.
15x³ + 20x² = 5x²(3x + 4)
[5x² is the common factor of each term]
Answers can be checked easily by expanding the brackets.

Always factorise completely. If the expression in the brackets has a common factor, then
the factorising is not complete.
Examples
6.
24x + 48 can be factorised as 2(12x + 24).
Since 12x and 24 have a common factor, factorising is not complete.
24x + 48 is completely factorised as 24(x + 2).
7.
r²h + 2rh can be factorised as
(r²h + 2rh),
r(rh + 2h) and
rh(r + 2).
The last expression has been completely factorised.
factorising
©RSH 24-Mar-10
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FACTORISING
Higher Tier
QUADRATICS OF THE FORM x² + bx + c

A quadratic expression contains an x² term.

These are quadratics:

Quadratic expressions are obtained when two brackets are multiplied together like this

When you factorise a quadratic, you must find the two brackets.

With practice, you can quickly find the terms needed.
x² + 5x  6, x²  6x + 8, x²  25
Examples
1. x² + 5x + 6
= (x
=(x
2. x² + 4x + 4
) (x
3) (x
)
The first term in each bracket is easy, it must
be x because only x multiplied by x equals x²
2)
Next think of all factors of 6 which add to give
the 5
= (x + 3) (x + 2)
Then put the correct signs in place
= (x
)
The first term in each bracket is easy, it must
be x because only x multiplied by x equals x²
2)
Next think of all factors of 4 which add to give
the 4
=(x
) (x
2) (x
= (x + 2) (x + 2)
Then put the correct signs in place
= (x + 2)²
factorising
©RSH 24-Mar-10
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FACTORISING
3. x² + 5x  6
4. x²  6x + 8
5. x²  25
Higher Tier
= (x
) (x
)
= (x  1) (x + 6)
Factors of 6 which add to give the 5 are 6
and 1
= (x
The first term in each bracket is easy, it must
be x because only x multiplied by x equals x²
) (x
)
= (x  4) (x  2)
Factors of 8 which add to give the 6 are 4
and 2
= (x
The first term in each bracket is easy, it must
be x because only x multiplied by x equals x²
) (x
)
= (x  5) (x + 5)
factorising
The first term in each bracket is easy, it must
be x because only x multiplied by x equals x²
Factors of 25 which add to give the 0 are 5
and 5
©RSH 24-Mar-10
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FACTORISING
Higher Tier
QUADRATICS OF THE FORM ax² + bx + c

We now look at quadratic expressions in which the first term is not 1x².

Two techniques are illustrated.
Examples
1. 3x² + 7x + 2
(3x
) (x
)
(3x + 2) (x + 1)
or
(3x + 1)(x + 2)
The first term in each bracket is easy,
because only 3x multiplied by x equals 3x²
There are two possibilities for the second
term in each bracket, because
2  1 = 2 and 1  2 = 2
Expand both and see which is correct
(3x + 2) (x + 1) = 3x² + 5x + 2
(3x + 1) (x + 2) = 3x² + 7x + 2
2. 5x²  7x + 2
= (3x + 1) (x + 2)
This is the correct factorisation
(5x
First term is easy – no choice
) (x
)
Possibilities
factorising
(5x + 2) (x  1)
No, this would give 2 as the last term
(5x  2) (x + 1)
No, this would also give 2
(5x  2) (x  1)
Yes, this gives 5x²  7x + 2
(5x + 1) (x  2)
No, this would also give 2
(5x  1) (x + 2)
No, this would also give 2
(5x + 1) (x + 2)
No, this gives 5x² +11x + 2
(5x + 2) (x + 1)
No, this gives 5x² +7x + 2
= (5x  2)(x  1)
Is the correct factorisation
©RSH 24-Mar-10
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FACTORISING
Higher Tier
Another useful technique follows:
3x² + 7x + 2






Consider the 3 and the 2.
3x2=6
Factors of 6 which add to give the 7 are:
Rewrite the expression as follows:
Use common factors:
Common factors again:
6 and 1
3x² + 6x + 1x + 2
3x(x + 2) + (x + 2)
(3x + 1)(x + 2)
6x²  5x  6






factorising
Consider the 6 and the 6.
6 x 6 = 36
Factors of 36 which add to give the 5 are:
Rewrite the expression as follows:
Use common factors:
Common factors again:
©RSH 24-Mar-10
9 and 4
6x²  9x + 4x  6
3x(2x  3) +2(2x  3)
(3x + 2)(2x  3)
Page 5 of 6
FACTORISING
Higher Tier
DIFFERENCE OF TWO SQUARES

You may have noticed that
(x + a)(x  a) = x²  a²
for example, (x + 4)(x  4) = x²  4x + 4x  4² = x²  4² = x²  16

This fact helps us to factorise the difference between two squares.
Examples
1.
2.
3.
4.
5.
x²  9
x²  25
4x²  b²
100  x²
81x²  16y²
factorising
=
x²  3²
=
(x + 3)(x  3)
=
x²  5²
=
(x + 5)(x  5)
=
(2x)²  b²
=
(2x + b)(2x  b)
=
10²  x²
=
(10 + x)(10  x)
=
(9x + 4y)(9x  4y)
©RSH 24-Mar-10
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