Lesson 6 ~35 min Unit 2 · Algebra +85 XP

Factorising Monic Quadratics

Every expression of the form $x^2 + bx + c$ hides a pair of brackets. Learn to hunt down the right factor pair, nail the signs, and verify every answer by expanding.

Today's hook: A rectangle has area $x^2 + 7x + 12$. Can you find its side lengths without drawing? The secret lies in finding two numbers that multiply to 12 and add to 7.

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Think First

warm-up

Before you read on — factorise $x^2 + 7x + 12$. What two numbers multiply to give $12$ and add to give $7$? Try it, then check your reasoning as you go.

Record your answer in your workbook.

The Big Idea

+5 XP to read

A monic quadratic has the form $x^2 + bx + c$. When you factorise it, you are looking for two numbers $m$ and $n$ that multiply to $c$ and add to $b$. Those two numbers become the constants inside the brackets: $(x + m)(x + n)$.

If $m \times n = c$ and $m + n = b$, then $x^2 + bx + c = (x + m)(x + n)$. The coefficient of $x^2$ is 1, so the brackets always start with $x$.

$x^2+bx+c=(x+m)(x+n)$

Monic means 1

The $x^2$ coefficient is 1. Non-monic quadratics (like $2x^2 + \dots$) need a different method.

Two conditions

The same pair must satisfy BOTH multiply-to-c and add-to-b. One condition is not enough.

Order free

$(x + 3)(x + 4)$ is the same as $(x + 4)(x + 3)$. The order of factors does not matter.

What You'll Master

objectives

Know

The form of a monic quadratic: $x^2 + bx + c$
How to list factor pairs of a number systematically
The sign rules for factorising monic quadratics

Understand

Why the two numbers must multiply to $c$ and add to $b$
Why $c$ positive means same signs, $c$ negative means opposite signs
How expanding reverses factorisation for verification

Can Do

Factorise monic quadratics with positive and negative constants
Apply sign rules confidently without guessing
Verify every answer by expanding the brackets

Words You Need

vocabulary

Monic quadraticA quadratic where the coefficient of $x^2$ is 1: $x^2 + bx + c$.

Factor pairTwo numbers that multiply to give a target value. For 12: $(1,12)$, $(2,6)$, $(3,4)$.

CoefficientThe number in front of a variable. In $x^2 + 7x + 12$, the coefficient of $x$ is 7.

Constant termThe term without a variable: $c$ in $x^2 + bx + c$.

Factorise fullyWrite as a product of brackets that cannot be factorised further.

ExpandMultiply out brackets to return to the original expression. Used to check factorisation.

Spot the Trap

heads-up

Wrong: "$x^2 - 7x + 12 = (x + 3)(x + 4)$" — signs are wrong. $c$ is positive and $b$ is negative, so both numbers must be negative.

Right: $x^2 - 7x + 12 = (x - 3)(x - 4)$ because $(-3) \times (-4) = 12$ and $(-3) + (-4) = -7$.

Wrong: "$x^2 + 2x - 8 = (x + 2)(x + 4)$" — wrong pair and wrong signs. Product would be $+8$, not $-8$.

Right: $x^2 + 2x - 8 = (x - 2)(x + 4)$ because $(-2) \times 4 = -8$ and $(-2) + 4 = 2$.

The Pair Hunt

+5 XP

To factorise $x^2 + bx + c$, you need to find the one pair of numbers that both multiplies to $c$ and adds to $b$. The secret is to list factor pairs systematically so you do not miss the right one.

For $x^2 + 7x + 12$: list pairs of 12: $(1,12)$, $(2,6)$, $(3,4)$. Check sums: $1+12=13$, $2+6=8$, $3+4=7$. The pair $(3,4)$ works, so $(x+3)(x+4)$.

$m \times n = c$
$m + n = b$

Start from 1

List pairs in order: $(1,c)$, $(2,c/2)$, $(3,c/3)$... Stop when you pass the square root of $c$.

Check every pair

Do not stop at the first pair you see. The correct pair might be further down your list.

Write it out

Showing your pair list earns method marks even if the final answer has a slip.

Same Sign

+5 XP

When $c$ is positive, the two numbers must have the same sign — both positive or both negative. Which one? Look at $b$: the numbers share the same sign as $b$.

$x^2 + 5x + 6$: $c=6$ is positive, $b=5$ is positive, so both numbers are positive. $2 \times 3 = 6$ and $2 + 3 = 5$. Result: $(x+2)(x+3)$.

$x^2-8x+15$
= (x-3)(x-5)

Positive c, same sign

Two negatives multiply to a positive. Two positives also multiply to a positive. Match $b$.

Follow b's lead

$b$ tells you which sign to use. If $b$ is negative, both numbers are negative.

Check the sum

$(-3) + (-5) = -8$. The sum is negative, matching $b = -8$.

Opposite Signs

+5 XP

When $c$ is negative, the two numbers must have opposite signs — one positive, one negative. The number with the larger absolute value takes the same sign as $b$.

$x^2 + 2x - 8$: $c=-8$ is negative, so mixed signs. $b=2$ is positive, so the larger absolute value is positive. $-2 \times 4 = -8$ and $-2 + 4 = 2$. Result: $(x-2)(x+4)$.

$x^2+2x-8$
= (x-2)(x+4)

Negative c = mixed

A negative product always requires one positive and one negative factor.

Larger abs wins

The number with the larger absolute value takes the same sign as $b$.

List both ways

For $-8$: try $(1,-8)$, $(-1,8)$, $(2,-4)$, $(-2,4)$. One of them will sum to $b$.

Check by Expanding

+5 XP

Always expand your answer to verify it matches the original. This catches sign slips, wrong factor pairs, and arithmetic errors before they cost you marks.

Check $(x + 3)(x + 6)$: First $= x^2$, Outer $= 6x$, Inner $= 3x$, Last $= 18$. Sum: $x^2 + 9x + 18$. Matches the original. Verified.

$(x+3)(x+6)$
= x² + 9x + 18

Catch sign slips

The most common error is getting the signs wrong. Expanding reveals this instantly.

Use FOIL

First, Outer, Inner, Last. Add the four products and check against the original.

5 seconds of safety

A quick expand takes five seconds and saves losing a whole mark.

Watch Me Solve It · 3 examples

Watch Me Solve It · Both numbers positive

+15 XP per step

PROBLEM

Factorise $x^2 + 9x + 18$.

1

Identify $b$ and $c$

$b = 9$ and $c = 18$

We need two numbers that multiply to 18 and add to 9.
2

List factor pairs of 18

$(1, 18)$, $(2, 9)$, $(3, 6)$

Systematic search from 1 upwards. Stop at $\sqrt{18} \approx 4.2$.
3

Check the sums

$1 + 18 = 19$, $2 + 9 = 11$, $3 + 6 = 9$

The pair $(3, 6)$ adds to 9. Both are positive, matching $b = 9$.
4

Write the factors and check

$x^2 + 9x + 18 = (x + 3)(x + 6)$

Check: $(x + 3)(x + 6) = x^2 + 6x + 3x + 18 = x^2 + 9x + 18$

Answer$(x + 3)(x + 6)$

Watch Me Solve It · Opposite signs

+15 XP per step

PROBLEM

Factorise $x^2 - 5x - 14$.

1

Identify $b$ and $c$

$b = -5$ and $c = -14$

$c$ is negative, so the numbers have opposite signs. $b$ is negative, so the larger absolute value is negative.
2

List mixed-sign factor pairs of $-14$

$(1, -14)$, $(-1, 14)$, $(2, -7)$, $(-2, 7)$

Both orderings: positive-then-negative and negative-then-positive.
3

Check the sums

$1 + (-14) = -13$, $(-1) + 14 = 13$, $2 + (-7) = -5$, $(-2) + 7 = 5$

The pair $(2, -7)$ sums to $-5$. The larger absolute value (7) is negative, matching $b$.
4

Write the factors and check

$x^2 - 5x - 14 = (x + 2)(x - 7)$

Check: $(x + 2)(x - 7) = x^2 - 7x + 2x - 14 = x^2 - 5x - 14$

Answer$(x + 2)(x - 7)$

Watch Me Solve It · Both numbers negative

+15 XP per step

PROBLEM

Factorise $x^2 - 8x + 15$.

1

Identify $b$ and $c$

$b = -8$ and $c = 15$

$c$ is positive, so both numbers have the same sign. $b$ is negative, so both are negative.
2

List negative factor pairs of 15

$(-1, -15)$, $(-3, -5)$

Only negative pairs are needed because both numbers must be negative.
3

Check the sums

$(-1) + (-15) = -16$, $(-3) + (-5) = -8$

The pair $(-3, -5)$ sums to $-8$, matching $b$.
4

Write the factors and check

$x^2 - 8x + 15 = (x - 3)(x - 5)$

Check: $(x - 3)(x - 5) = x^2 - 5x - 3x + 15 = x^2 - 8x + 15$

Answer$(x - 3)(x - 5)$

Common Pitfalls

heads-up

Wrong signs

Using $(x + 3)(x + 4)$ for $x^2 - 7x + 12$. If $c$ is positive and $b$ is negative, both numbers must be negative.

Fix: Check the sign of $c$ first. Positive $c$ means same sign as $b$.

Wrong factor pair

Choosing $(2, 6)$ for $c = 12$ when the sum needed is 7, not 8. Students often stop at the first pair they see.

Fix: List ALL factor pairs systematically. Check every sum.

Forgetting to check by expanding

Writing an answer and moving on without verification. A quick expand catches most errors instantly.

Fix: Make expanding a habit. Five seconds of checking saves lost marks.

Copy Into Your Books

The Method

For $x^2 + bx + c$, find $m$ and $n$ where $m \times n = c$ and $m + n = b$
Write: $(x + m)(x + n)$
Always check by expanding

Sign Rules

$c$ positive: same sign as $b$
$c$ negative: opposite signs; larger abs matches $b$
When in doubt, list all pairs

Systematic Search

Start from 1, list pairs in order
Stop when you pass $\sqrt{c}$
Include negative pairs when $c$ is negative

What Not to Do

Do not guess — list pairs
Do not ignore signs
Do not skip the expansion check

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Mixed

4 problems

Four problems covering positive $c$, negative $c$, and mixed sign cases. Work each one, then reveal the answer to check.

1 Factorise $x^2 + 5x + 6$.

Need numbers that multiply to 6 and add to 5$= (x + 2)(x + 3)$
2 Factorise $x^2 - 7x + 12$.

$c$ positive, $b$ negative: both numbers negative. $(-3) \times (-4) = 12$, $(-3) + (-4) = -7$$= (x - 3)(x - 4)$
3 Factorise $x^2 + 2x - 15$.

$c$ negative: opposite signs. Larger abs positive (matches $b = 2$). $-3 \times 5 = -15$, $-3 + 5 = 2$$= (x - 3)(x + 5)$
4 Factorise $x^2 - 4x - 21$.

$c$ negative: opposite signs. Larger abs negative (matches $b = -4$). $3 \times (-7) = -21$, $3 + (-7) = -4$$= (x + 3)(x - 7)$

Complete in your workbook.

Quick Check · 5 questions

Which pair of numbers multiplies to $12$ and adds to $7$?

+10 XP

Factorise $x^2 - 6x + 8$.

+10 XP

For $x^2 + 3x - 10$, what signs must the two factors have?

+10 XP

Factorise $x^2 + x - 20$.

+10 XP

When factorising $x^2 - 9x + 20$, which of the following are common errors?

+10 XP

Show Your Working · 3 questions

Show Your Working

7 marks total

Apply Easy 2 MARKS

Q6. Factorise $x^2 + 11x + 24$.

Answer in your workbook.

Analyse Medium 2 MARKS

Q7. The quadratic $x^2 + bx + 12$ can be factorised as $(x + 2)(x + 6)$. Find the value of $b$.

Answer in your workbook.

Evaluate Hard 3 MARKS

Q8. Explain the steps you would use to factorise $x^2 - 3x - 18$. In your answer, include why the signs of the two numbers must be different.

Answer in your workbook.

Comprehensive Answers

Quick Check

1. B — $3 \times 4 = 12$ and $3 + 4 = 7$.

2. C — $c = 8$ positive, $b = -6$ negative, so both numbers are negative. $(-2) \times (-4) = 8$ and $(-2) + (-4) = -6$.

3. C — $c = -10$ is negative, so the numbers must have opposite signs.

4. B — Need numbers that multiply to $-20$ and add to $1$. The pair $-4$ and $5$ works.

5. C — Both A and B are errors. We need factors of 20 that add to $-9$, and both numbers must be negative.

Show Your Working Model Answers

Q6 (2 marks): Factor pairs of 24: $(1,24)$, $(2,12)$, $(3,8)$, $(4,6)$ [1]. Sums: $1+24=25$, $2+12=14$, $3+8=11$, $4+6=10$. None add to 11... wait, $3+8=11$ [1]. So $(x + 3)(x + 8)$.

Q7 (2 marks): Method: $b = 2 + 6$ (or expanding $(x+2)(x+6)$) [1]. $b = 8$ [1].

Q8 (3 marks): Step 1: Identify $b = -3$ and $c = -18$ [1]. Step 2: Since $c = -18$ is negative, the two numbers must have opposite signs (a negative product requires one positive and one negative factor) [1]. Step 3: Find pair that multiplies to $-18$ and adds to $-3$: $3$ and $-6$ [1]. So $(x + 3)(x - 6)$.

Stretch Challenge · +25 XP, +10 coins

Factorise Twice

Fully factorise $x^4 - 13x^2 + 36$. Hint: let $u = x^2$, factorise the quadratic in $u$, then substitute back. Watch out — one of your factors might be factorisable again.

Reveal solution

Let $u = x^2$. Then $x^4 - 13x^2 + 36 = u^2 - 13u + 36$.

Factorise: need two numbers that multiply to 36 and add to $-13$. Those are $-4$ and $-9$.

So $u^2 - 13u + 36 = (u - 4)(u - 9) = (x^2 - 4)(x^2 - 9)$.

Now $x^2 - 4 = (x + 2)(x - 2)$ and $x^2 - 9 = (x + 3)(x - 3)$.

Fully factorised: $(x + 2)(x - 2)(x + 3)(x - 3)$.

Quick Review

The Pair Hunt

Find $m$ and $n$ where $m \times n = c$ and $m + n = b$

$c$ Positive

Both numbers share the sign of $b$

$c$ Negative

Opposite signs; larger abs matches $b$

Systematic Search

List all pairs, check every sum

Check by Expanding

Verify with FOIL every time

Sign Errors

Match the sign of $c$ to the rule

Interactive: Quadratic Factoriser

Explore factor pairs for any monic quadratic, then practise factorising with instant feedback.

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