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Solving Quadratics
A quadratic equation is one where the highest power is 2. There are 3 ways of solving
them: factorising, using the formula and completing the square:
This lesson will look at the method of factorisation. (See the previous lesson for how to
use the formula or complete the square.)
Factorising when the coefficient of 𝒙𝟐 is 1:
Factorising a quadratic usually means putting it into 2 brackets (i.e. the reverse of
multiplying out brackets seen in lesson 3).
For example:
Factorises to become:
𝒙𝟐 + 𝟐𝒙 − 𝟏𝟓
(𝒙 + 𝟓)(𝒙 − 𝟑)
This is the same expression
factorised into 2 brackets.
)(𝑥
To factorise an expression, write 2 brackets with an 𝑥 in each [(𝑥
)] then follow
the systematic steps to work out the correct signs and numbers to go in the brackets:

1st look at the sign of the number on its own:
+𝒗𝒆 number means the signs in the brackets will be the same: (𝑥 + )(𝑥+ ) or (𝑥 − )(𝑥− )
−𝒗𝒆 number means that the signs in the brackets will be different: (𝑥 + )(𝑥− )
Then follow the correct route below:
If Same signs:


e.g.(𝑥 + )(𝑥+ ) or (𝑥 − )(𝑥− )
Look at the coefficient of 𝒙:
If it’s +ve then the signs in the brackets are +ve
if it’s –ve then the signs in the brackets are –ve
(𝑥 + )(𝑥+ )
(𝑥 − )(𝑥− )
Finally find the numbers for each bracket:
The 2 numbers need to multiply to give the number on its own, and add to give the
coefficient of 𝑥.
If Different signs: e.g. (𝑥 + )(𝑥− )

Find the numbers for each bracket:
The 2 numbers need to multiply to give the number on its own, and have a
difference of the coefficient of 𝑥.

Look at the coefficient of 𝒙:
If it’s +ve then put the biggest number in the bracket with the + sign.
if it’s –ve then put the biggest number in the bracket with the − sign.
© H Jackson 2012 / ACADEMIC SKILLS
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Examples:

Factorise: 𝒙𝟐 + 𝟕𝒙 + 𝟏𝟐
Step 1:
Step 2:
Step 3:
Step 4:
we have +𝟏𝟐 so the signs must be the same.
we have +𝟕𝒙 so the signs must both be positive.
so far we have, (𝒙 + )(𝒙+ )
work out the numbers (3x4=12 and 3+4=7)
Answer:
(𝑥 + 4)(𝑥 + 3)

Factorise: 𝒙𝟐 − 𝟗𝒙 + 𝟐𝟎
Step 1:
Step 2:
Step 3:
Step 4:
we have +𝟐𝟎 so the signs must be the same.
we have −𝟗𝒙 so the signs must both be negative.
so far we have, (𝒙 − )(𝒙− )
work out the numbers (4x5=20 and 4+5=9)
Answer:
(𝑥 − 4)(𝑥 − 5)

Factorise: 𝒙𝟐 + 𝟑𝒙 − 𝟒𝟎
Step 1:
Step 2:
Step 3:
Step 4:
we have −𝟒𝟎 so the signs must be the different.
so far we have, (𝒙 + )(𝒙− )
work out the numbers (8x5=40 and 8-5=3)
we have +𝟑𝒙 so the biggest number goes with the + sign.
Answer:
(𝑥 + 8)(𝑥 − 3)

Factorise: 𝒙𝟐 − 𝟑𝒙 − 𝟏𝟖
Step 1:
Step 2:
Step 3:
Step 4:
we have −𝟏𝟖 so the signs must be the different.
so far we have, (𝒙 + )(𝒙− )
work out the numbers (6x3=18 and 6-3=3)
we have −𝟑𝒙 so the biggest number goes with the - sign.
Answer:
(𝑥 + 3)(𝑥 − 6)
(n.b. once you have the 2 brackets you can multiply them out to check that you get what
you started with.)
© H Jackson 2012 / ACADEMIC SKILLS
2
Factorising when the coefficient of 𝒙𝟐 is not 1:
If the coefficient of 𝑥 2 is not 1 you can still use the above method but you would have to
use trial and error to decide what the coefficients of 𝑥 should be in the brackets. The
following shows a systematic method for factorising these types of quadratics:
Examples:
 Factorise 𝟔 𝒙𝟐 − 𝟏𝟏𝒙 − 𝟏𝟎
𝟔 𝒙𝟐 − 𝟏𝟏𝒙 − 𝟏𝟎
Step 1:
do 6 × −10 = −60
Step 2:
find 2 numbers that multiply to −60 and add or subtract to −11.
Step 3:
±1 × ∓60 = −60, can’t make −11
X
±2 × ∓30 = −60, can’t make −11
X
±3 × ∓20 = −60, can’t make −11
X
±4 × ∓15 = −60, −15 + 4 = −11

(±5 × ∓12 = −60, can’t make −11
X)
(±6 × ∓10 = −60, can’t make −11
X)
Split the middle value (using the numbers you found) and rewrite the
equation:
(Notice this is the same
equation written differently:
−15𝑥 + 4𝑥 = −11𝑥)
𝟔 𝒙𝟐 − 𝟏𝟓𝒙 + 𝟒𝒙 − 𝟏𝟎
Step 4:
Split the equation into 2 and factorise each half:
𝟔 𝒙𝟐 − 𝟏𝟓𝒙 + 𝟒𝒙 − 𝟏𝟎
3𝑥(2𝑥 − 5)
2(2𝑥 − 5)
(n.b. the bit in the bracket should always be exactly the same for each half)
Step 5:
Write the 2 brackets: (3𝑥 + 2)(2𝑥 − 5)
(Notice the (3𝑥 + 2) comes from the bits outside the brackets)
 Factorise: 𝟖 𝒙𝟐 + 𝟏𝟎𝒙 + 𝟑
Step 3:
8 × 3 = 24
find 2 numbers that multiply to 24 and add or subtract to 10.
6 × 4 = 24, 6 + 4 = 10

(n.b. if there’s nothing else to
2
8 𝑥 + 6𝑥 + 4𝑥 + 3
Step 4:
2𝑥(4𝑥 + 3)
Step 5:
Write the 2 brackets: (2𝑥 + 1)(4𝑥 + 3) = 0
Step 1:
Step 2:
+ 1(4𝑥 + 3)
factorise you need to put a 1
outside the bracket.)
(n.b. if you prefer just to learn one method, the above can also be used when the coefficient of 𝑥 2 =1)
© H Jackson 2012 / ACADEMIC SKILLS
3
Factorising can be used to solve a quadratic (i.e. find the values of 𝑥 that make it equal
zero). The values that you find are called the “roots” of the equation – if you sketched the
graph these points would be where the graph crosses the 𝑥-axis.
Solving quadratics:
The quadratic must = 0 (if it doesn’t then rearrange it).
Remember: If 2 things multiply together to give 0, then at least one of them must =0.
We use this fact to solve a quadratic once we have factorised it.
Let’s look at the 6 examples that we already factorised, put them =0 and solve them.
Examples:

Solve the following by factorising:
Factorise to give:
𝒙𝟐 + 𝟕𝒙 + 𝟏𝟐 = 𝟎
(𝑥 + 4)(𝑥 + 3) = 0
(Bracket x bracket = 0, so one of the brackets must be 0.)
Either:
Rearrange:
(𝑥 + 4) = 0 or:
𝑥 = −4
or:

𝒙𝟐 − 𝟗𝒙 + 𝟐𝟎 = 𝟎
Solve the following by factorising:
Factorise to give:
Either:
(𝑥 + 3) = 0
𝑥 = −3
(𝑥 − 4)(𝑥 − 5) = 0
(𝑥 − 4) = 0 or: (𝑥 − 5) = 0
∴ 𝑥 = 4 𝑜𝑟 𝑥 = 5

Solve the following by factorising:
Factorise to give:
Either:
𝒙𝟐 + 𝟑𝒙 − 𝟒𝟎 = 𝟎
(𝑥 + 8)(𝑥 − 5) = 0
(𝑥 + 8) = 0 or: (𝑥 − 5) = 0
∴ 𝑥 = −8 𝑜𝑟 𝑥 = 5

Solve the following by factorising:
Factorise to give:
Either:
𝒙𝟐 − 𝟑𝒙 − 𝟏𝟖 = 𝟎
(𝑥 + 3)(𝑥 − 6) = 0
(𝑥 + 3) = 0 or: (𝑥 − 6) = 0
∴ 𝑥 = −3 𝑜𝑟 𝑥 = 6
© H Jackson 2012 / ACADEMIC SKILLS
4
 Solve the following by factorising:
𝟔 𝒙𝟐 − 𝟏𝟏𝒙 − 𝟏𝟎 = 𝟎
Factorise to give:
Either:
(3𝑥 + 2)(2𝑥 − 5) = 0
(3𝑥 + 2) = 0 or:
(2𝑥 − 5) = 0
2
5
3
2
∴ 𝑥 = − 𝑜𝑟 𝑥 =
 Solve the following by factorising:
𝟖 𝒙𝟐 + 𝟏𝟎𝒙 + 𝟑 = 𝟎
Factorise to give:
Either:
(2𝑥 + 1)(4𝑥 + 3) = 0
(2𝑥 + 1) = 0 or:
(4𝑥 + 3) = 0
1
3
2
4
∴ 𝑥 = − 𝑜𝑟 𝑥 = −
One more thing:

Solve: −𝒙𝟐 − 𝟕𝒙 + 𝟏𝟐 = 𝟎
This looks a bit more tricky because we have −𝑥 2 . Probably the easiest way to solve this
is to multiply everything by −1 (i.e. change all the signs) and then you can factorise as
normal.
Now we have: 𝒙𝟐 + 𝟕𝒙 − 𝟏𝟐 = 𝟎
(try it and check that your answers work)
(we don’t need to put −0 as
it’s the same thing.)
© H Jackson 2012 / ACADEMIC SKILLS
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