© Daniel Holloway
Factorisation of quadratics involves putting
quadratic expressions back into brackets (if
possible). To do this, the expression can be
in the form:
x2 + ax + b
Where a and b are integers
We can tell how to start off the brackets by
looking at the signs of the expression:
 x2 + ax + b = (x + ?)(x + ?)
because everything is positive
 x2 - ax + b = (x - ?)(x - ?)
because negative x negative = positive


When the second sign is a plus, both
brackets are the same as the first sign
x2 + ax + b = (x + ?)(x + ?)
x2 - ax + b = (x - ?)(x - ?)
When the second sign is a minus, both
brackets are different signs
x2 + ax - b = (x + ?)(x - ?)
x2 - ax - b = (x + ?)(x - ?)
When a quadratic expression is in the form
x2 + ax + b, we must make sure that the
numbers in the brackets satisfy these rules:
 the b must be the product of the two
numbers
 the a must be the sum of the numbers
when the two brackets are the same sign
 the a must be the difference between
the numbers when the two brackets are
different signs
Take this example:
x2 + 5x + 6
Both signs are positive, so we can begin with
(x + ?)(x + ?)
The numbers need to multiply to make 6 and
add together to make 5
They must be 2 and 3
Since both signs are positive, the order does
not matter:
(x + 2)(x + 3)
Take this example:
y2 – y – 90
Both signs are negative, so we can begin
with (x + ?)(x - ?)
The numbers need to multiply to make 90
and add together to make 1
They must be 9 and 10
Because we need to do 9 - 10 to get -1:
(y + 9)(y - 10)
1.
Factorise the following…
a) x2 + 5x + 4
b) x2 + 7x + 10
c) y2 + 14y + 24
d) z2 + 9z + 18
e) a2 - 6a - 7
f) x2 - x - 12
g) k2 + 4k + 3
h) s2 - 18c + 32
2.
Factorise x2 – 24x + 144
As a rule, (a + b)(a – b) multiplies out to get a2 b2
This type of quadratic expression where there
are only two parts to it, both of which are
perfect squares, separated by a minus sign, is
called the difference of two squares. Examples
include:
a2 - 9
a2 - 25
a2 – 100
To factorise them, we simply root them and
put them in the format shown above.
e.g. a2 - 9 factorises to (a + 3)(a - 3)
We can solve quadratic equations in the
form x2 + ax + b by first factorising it, and
finding the two answers using the double
brackets.
e.g. Solve x2 + 6x + 5 = 0
This factorises into (x + 5)(x + 1)
If x + 5 = 0, then x = -5
If x + 1 = 0, then x = -1
We use the same method of factorisation
to factorise quadratic expressions in the
form ax2 + bx + c but we have to take into
account the a which will be in a bracket
The following slide explains how it’s done
Factorise 3x2 + 8x + 4. The same bracket rules
apply, so both brackets are positive.
As 3 only has the factors 3 and 1:
(3x + ?)(x + ?)
Next we notice the factors of 4 are 1x4 and
2x2. The only pair here which combine with
3x1 to make 8 is 2x2:
(3 x 2) + (1 x 2) = 8
So the factorised expression is:
(3x + 2)(x + 2)
There are THREE ways we can solve
quadratic equations in the form ax2 + bx + c
as shown on the next few slides.
Solve 12x2 - 28x = -15.
1. Rearrange the equation to equal zero
12x2 - 28x + 15 = 0
2. Factorise the expression
(2x - 3)(6x - 5)
3. Find the two answers as before
x = 1.5 and x = 25/6
Many quadratic equations cannot be
factorised because they are too
complicated or do not have convenient
integer factors between them. This is when
the quadratic formula is used:
-b ± √b2 - 4ac
x=
2a
Solve 5x2 - 11x – 4 = 0.
1. Remember the quadratic formula:
2.
3.
-b ± √b2 - 4ac
x=
2a
Substitute the values from the equation
as a, b and c
Solve the equation using the values,
shown on the following slide
Solve 5x2 - 11x – 4 = 0.
We know that a = 5, b = -11 and c = -4
-b ± √b2 - 4ac
x=
2a
11 ± √201
=
10
11 ± √121 – 4(5)(-4)
x=
10
x = 2.52
x = 0.32
Remember that (x + a)2 = x2 + 2ax + a2
This means that x2 + 2ax = (x + a)2 - a2
This is the principle behind completing the
square, a rule used to solve quadratic
equations
There are two steps to completing the
square, as shown on the following slide
Step 1:
Rewrite x2 + 4x – 7 in the form (x + a)2 – b
Note that
x2 + 4x = (x + 2)2 – 4
Therefore
x2 + 4x – 7 = (x + 2)2 – 4 – 7
= (x + 2)2 – 11
Step 2:
Hence solve the equation x2 + 4x – 7 = 0
We have:
x2 + 4x –7 = (x + 2)2 – 11
When x2 + 4x – 7 = 0, we know:
(x + 2)2 – 11 = 0
(x + 2)2 = 11
Now square root both sides:
x + 2 = ±√11
x = -2 ±√11
So, x = 1.32 and x = 5.32 (to 2 d.p.)