Monic and non-monic factorisation | Year 10

01 / LEARN

The idea, step by step

Start with the simplest factor

First remove any factor common to every term. For a monic quadratic, look for two numbers whose product is the constant term and whose sum is the coefficient of x.

A negative product needs opposite signs. A positive product needs matching signs; the required sum tells you which sign.

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

When the leading coefficient is not one

For a non-monic quadratic, the split-middle-term method keeps the arithmetic organised. Find two numbers with product ac and sum b. Split bx using them, then factor each pair.

The two groups must expose the same bracket. Expand the finished answer to check every coefficient.

a x^{2} + b x + c = a x^{2} + p x + q x + c, p + q = b, p q = a c

A common trap

Choosing numbers that multiply to b rather than to ac.

Instead: For the split-middle-term method, the product target includes both the leading coefficient and the constant.

02 / WORKED EXAMPLES

See the thinking

Example 1

Find a product and sum

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

The pair $- 4$ and $3$ has product $- 12$ and sum $- 1$ .
Use these numbers in the brackets.
Expanding $(x - 4) (x + 3)$ gives $x^{2} - x - 12$ .

(x - 4) (x + 3)

Example 2

Split a non-monic middle term

Factorise $6 x^{2} + 7 x - 3$ .

$a c = - 18$ ; the required numbers are $9$ and $- 2$ .
$6 x^{2} + 9 x - 2 x - 3 = 3 x (2 x + 3) - 1 (2 x + 3)$ .
Factor out the common bracket.

(3 x - 1) (2 x + 3)

Check your understanding

Factorise $x^{2} + 7 x + 12$ .

Show answer

$(x + 3) (x + 4)$

$3 \times 4 = 12$ and $3 + 4 = 7$ .

03 / WORKSHEET

Your turn

6 questions · 16 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Factorise $x^{2} + 9 x + 20$ .
2 marks
Factorise $x^{2} - 2 x - 15$ .
2 marks
Factorise $2 x^{2} + 7 x + 3$ .
3 marks
Factorise $3 x^{2} - 11 x - 4$ .
3 marks
Factorise $2 x^{2} - 18$ completely.
3 marks
Can $x^{2} + x + 1$ be written as $(x + p) (x + q)$ with integer p and q? Explain.
3 marks

Read the worked solutions

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

Factorise $x^{2} + 9 x + 20$ .
2 marks
$4 \times 5 = 20$ and $4 + 5 = 9$ .
$(x + 4) (x + 5)$
Factorise $x^{2} - 2 x - 15$ .
2 marks
$- 5 \times 3 = - 15$ and $- 5 + 3 = - 2$ .
$(x - 5) (x + 3)$
Factorise $2 x^{2} + 7 x + 3$ .
3 marks
Split $7 x$ as $6 x + x$ : $2 x (x + 3) + (x + 3)$ .
$(2 x + 1) (x + 3)$
Factorise $3 x^{2} - 11 x - 4$ .
3 marks
Split $- 11 x$ as $x - 12 x$ : $x (3 x + 1) - 4 (3 x + 1)$ .
$(3 x + 1) (x - 4)$
Factorise $2 x^{2} - 18$ completely.
3 marks
First remove $2$ : $2 (x^{2} - 9)$ .
Then use the difference of two squares.
$2 (x - 3) (x + 3)$
Can $x^{2} + x + 1$ be written as $(x + p) (x + q)$ with integer p and q? Explain.
3 marks
The integer factor pairs of $1$ are $(1, 1)$ and $(- 1, - 1)$ . Their sums are $2$ and $- 2$ , not $1$ .
No integer factorisation.

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Curriculum reference & lesson scope

Australian Curriculum v9.0 and South Australian Curriculum

Year 10 quadratics, including the fuller school-course algebra requested for Ascent: non-monic factorisation, completing the square and the quadratic formula. These lessons consolidate Year 9 monic quadratics and extend the available solving methods.

Official curriculum source ↗

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