Factorising monic quadratics | Year 9

01 / LEARN

The idea, step by step

Read both clues

A monic quadratic has coefficient $1$ on $x^{2}$ . In $x^{2} + b x + c$ , a factorisation $(x + p) (x + q)$ needs $p + q = b$ and $p q = c$ . The middle coefficient gives a sum and the constant gives a product. List factor pairs of $c$ and test their sums.

For $x^{2} + 7 x + 12$ , the positive factor pairs of $12$ are $(1, 12)$ , $(2, 6)$ and $(3, 4)$ . Only $3 + 4 = 7$ , so the factors are $(x + 3) (x + 4)$ . Expansion is the check.

Use signs before searching

If $c$ is positive, the two numbers have the same sign. A positive sum means both are positive; a negative sum means both are negative. If $c$ is negative, the signs differ. The number with larger magnitude determines the sign of the sum.

Not every quadratic has integer factors. If no integer factor pair has the needed sum, do not force a pair that only fits the product. This lesson focuses on expressions that do have integer factors.

A common trap

Factor pairs only need to multiply to the constant.

Instead: They must also add to the coefficient of $x$ ; use both conditions.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example

Factorise $x^{2} - 9 x + 20$ .

The product is $20$ and the sum is $- 9$ .
Both numbers must be negative: $- 4$ and $- 5$ .
Check: $x^{2} - 5 x - 4 x + 20 = x^{2} - 9 x + 20$ .

(x - 4) (x - 5)

Example 2

Worked example

Factorise $x^{2} + x - 12$ .

A negative product needs opposite signs.
$4 \times (- 3) = - 12$ and $4 + (- 3) = 1$ .
Expanding confirms the middle coefficient is $1$ .

(x + 4) (x - 3)

Check your understanding

Factorise $x^{2} + 8 x + 15$ .

Show answer

$(x + 3) (x + 5)$

$3 + 5 = 8$ and $3 \times 5 = 15$ .

03 / WORKSHEET

Your turn

6 questions · 13 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Factorise $x^{2} + 5 x + 6$ .
2 marks
Factorise $x^{2} - 8 x + 15$ .
2 marks
Factorise $x^{2} + 2 x - 15$ .
2 marks
Factorise $x^{2} - 4 x - 21$ .
2 marks
Factorise $x^{2} - 16$ .
2 marks
Alex writes $x^{2} + 7 x + 10 = (x + 1) (x + 10)$ . Explain the error and correct it.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Factorise $x^{2} + 5 x + 6$ .
2 marks
$2 + 3 = 5$ and $2 \times 3 = 6$ .
$(x + 2) (x + 3)$
Factorise $x^{2} - 8 x + 15$ .
2 marks
$- 3 - 5 = - 8$ and $(- 3) (- 5) = 15$ .
$(x - 3) (x - 5)$
Factorise $x^{2} + 2 x - 15$ .
2 marks
$5 - 3 = 2$ and $5 (- 3) = - 15$ .
$(x + 5) (x - 3)$
Factorise $x^{2} - 4 x - 21$ .
2 marks
$- 7 + 3 = - 4$ and $- 7 (3) = - 21$ .
$(x - 7) (x + 3)$
Factorise $x^{2} - 16$ .
2 marks
The missing $x$ coefficient is zero.
$- 4 + 4 = 0$ and $(- 4) (4) = - 16$ .
$(x - 4) (x + 4)$
Alex writes $x^{2} + 7 x + 10 = (x + 1) (x + 10)$ . Explain the error and correct it.
3 marks
$1 + 10 = 11$ , not $7$ .
$2 + 5 = 7$ and $2 \times 5 = 10$ .
$(x + 2) (x + 5)$

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Curriculum reference & lesson scope

Australian Curriculum v9 / SA Curriculum Mathematics

AC9M9A02, AC9M9A04–A06: monic quadratic expressions, integer-root equations, graphs and models.

Official curriculum source ↗

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