Year 9AC9M9A02

Factorising Trinomials

Expand and factorise algebraic expressions, including monic and simple non-monic quadratic expressions.

The multiply-a-by-c method. It works for every trinomial, monic or not — and once you see why, you will never forget the steps.

Builds onYr 8 · Expanding brackets (FOIL)Yr 8 · Factors and factor pairsYr 7 · Integers: multiplying negatives

Why this works

Factorising is just expanding, run backwards. Start with what expanding two brackets actually produces:

Expand a general pair of brackets

(x+p)(x+q) = x^{2} + (p+q)x + pq

So for a monic trinomial

x^{2}+bx+c

, you need two numbers that add to $b$ and multiply to $c$ . That is the whole monic method.

Non-monic looks harder, but multiply through by

a

and watch what happens

a\left(ax^{2}+bx+c\right) = (ax)^{2} + b(ax) + ac

Let

u = ax

. The right-hand side becomes monic again

u^{2} + bu + ac

That is why you multiply $a$ by $c$ — you are turning a non-monic trinomial into a monic one in the variable $u = ax$ . And because you multiplied by $a$ at the start, you must divide it back out at the end. That is exactly what step 5 is doing. The method is not a trick; it is the monic method wearing a disguise.

Worked example

e.g.

3x^{2} + 20x - 7

Make sure it is written in the form $ax^{2}+bx+c$ $3x^{2} + 20x - 7$
Take out a common factor first if possible — here there is none
Multiply $a$ by $c$ $3 \times (-7) = -21$
Find the factor pair of that answer which adds to $b$ $+21 \text{ and } -1 \quad (21 - 1 = 20 = b \ \checkmark)$
Divide both factors by $a$ and simplify — this undoes the multiplication from the 'why' above $\frac{21}{3} = \frac{7}{1} \qquad \frac{-1}{3} \text{ (will not simplify)}$
Substitute the pair back into brackets: numerator is the constant, denominator is the coefficient of $x$ $(x+7)(3x-1)$
Always check by expanding. This catches almost every error you can make $(x+7)(3x-1) = 3x^{2} - x + 21x - 7 = 3x^{2} + 20x - 7 \ \checkmark$

Common mistakes

Wrong

x^{2} - 9x + 8 = (x-8)(x+1)

Right

x^{2} - 9x + 8 = (x-8)(x-1)

The classic sign error. Both numbers must multiply to

+8

and add to

-9

. With

-8

and

+1

you get

-8

and

-7

— neither is right. Expanding the wrong answer gives

x^{2}-7x-8

, which catches it instantly.

Wrong

2x^{2} + 10x + 12 \rightarrow ac = 24 \ldots

Right

2x^{2} + 10x + 12 = 2\left(x^{2}+5x+6\right) = 2(x+2)(x+3)

Skipping step 2. The

a \times c

method still works here, but taking out the common factor of 2 first turns a fiddly non-monic problem into an easy monic one. Always look for it.

Wrong

x^{2} + 4x + 4 = (x+?)(x+?) \text{ — hunting for two different numbers}

Right

x^{2} + 4x + 4 = (x+2)^{2}

Nothing says the two numbers must differ. Here both are

+2

. Perfect-square trinomials are common and are worth learning to spot: the constant is a perfect square and

b

is twice its root.

Wrong

\text{Assuming every trinomial factorises}

Right

\text{Check the discriminant: } b^{2}-4ac

Most do not. If

b^{2}-4ac

is not a perfect square, the trinomial has no integer factorisation and no amount of searching will find one. Checking first saves a lot of wasted effort — see Reasoning question 1.

Practice

FluencyGet quick and accurate at the method.

x^{2} + x - 20

Answer

(x+5)(x-4)

x^{2} - 9x + 8

Answer

(x-8)(x-1)

x^{2} + 6x + 8

Answer

(x+4)(x+2)

x^{2} + 11x + 24

Answer

(x+8)(x+3)

x^{2} + 4x + 4

Answer

(x+2)^{2}

9x^{2} + 12x + 4

Answer

(3x+2)^{2}

x^{2} + 2x - 3

Answer

(x+3)(x-1)

x^{2} + 8x + 15

Answer

(x+5)(x+3)

2x^{2} + 10x + 12

Answer

2(x+2)(x+3)

3x^{2} + 20x - 7

Answer

(x+7)(3x-1)

ReasoningExplain why. Say it in your own words.

Without factorising any of them, decide which of these cannot be factorised over the integers, and explain how you can tell:

x^{2}+4x+4

\ x^{2}-12x-36

\ x^{2}+8x+15

Answer

x^{2}-12x-36

cannot. Check the discriminant

b^{2}-4ac

: for the first,

16-16=0

(perfect square, factorises as

(x+2)^{2}

); for the third,

64-60=4

(perfect square); for the middle one,

144+144=288

, which is not a perfect square, so it has no integer factorisation. Its roots are

6 \pm 6\sqrt{2}

Explain why you should take out a common factor before using the

a \times c

method rather than after.

AnswerTaking it out first usually reduces the trinomial to monic, where you only need factors of

c

adding to

b

— no dividing by

a

at the end. Doing it afterwards still works but forces you through larger numbers. For

2x^{2}+10x+12

, factoring first gives

ac=6

instead of

ac=24

Someone writes

x^{2}-9x+8 = (x-8)(x+1)

. Identify the error, and describe a check that would have caught it in one line.

AnswerThe signs are wrong:

-8 \times +1 = -8

, not

+8

. Expanding gives

x^{2}-7x-8 \neq x^{2}-9x+8

. Expanding the answer back out is the check — it takes seconds and catches essentially every factorising error.

x^{2}+bx+c

, suppose

c>0

and

b<0

. What can you say about the signs of the two numbers in the brackets? Justify your answer.

AnswerBoth must be negative. Their product

c

is positive, so they share a sign; their sum

b

is negative, so that shared sign must be negative. This lets you fix the signs before you start searching.

Show that factorising

2x^{2}+7x+3

by the

a \times c

method gives the same result as multiplying through by

a

, factorising as a monic in

u=2x

, then dividing by

a

AnswerVia

a \times c

ac=6

, pair

6

and

1

, divide by 2 gives

\tfrac{6}{2}=\tfrac{3}{1}

and

\tfrac{1}{2}

, so

(x+3)(2x+1)

. Via substitution:

2(2x^{2}+7x+3) = u^{2}+7u+6 = (u+6)(u+1)

with

u=2x

, giving

(2x+6)(2x+1) = 2(x+3)(2x+1)

; dividing by 2 gives

(x+3)(2x+1)

. Identical — which is the point of the 'why' section.

AppliedThe same maths, inside a real question.

A rectangular courtyard has area

x^{2}+7x+12

square metres. Write expressions for its length and width.

Answer

(x+4)

m and

(x+3)

m. Area factorises as

(x+4)(x+3)

, and the two factors are the side lengths.

A ball is thrown so its height after

t

seconds is

h = -t^{2}+5t+6

metres. When does it hit the ground?

AnswerAfter 6 seconds. Set

h=0

-t^{2}+5t+6=0

, so

t^{2}-5t-6=0

, giving

(t-6)(t+1)=0

and

t=6

t=-1

. Reject

t=-1

— negative time has no meaning here.

A garden bed measures 8 m by 12 m. A path of uniform width

w

is laid around it, bringing the total area to 192 m². Find

w

Answer

w = 2

m. Total area

(8+2w)(12+2w)=192

, so

4w^{2}+40w+96=192

, giving

w^{2}+10w-24=0

and

(w+12)(w-2)=0

. Reject

w=-12

— a width cannot be negative.

Two consecutive positive integers have a product of 132. Find them.

Answer11 and 12. Let them be

n

and

n+1

n^{2}+n-132=0

, so

(n-11)(n+12)=0

and

n=11

(rejecting

n=-12

as it is not positive).

A photograph measures

(x+3)

cm by

(x+5)

cm and has an area of 99 cm². Find its dimensions.

Answer9 cm by 11 cm. Expanding gives

x^{2}+8x+15=99

, so

x^{2}+8x-84=0

and

(x-6)(x+14)=0

, giving

x=6

. Then the sides are

6+3=9

and

6+5=11

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