Monic vs Non-monic Quadratics (Year 10)

01 / LEARN

Understand the idea.

Start with the simplest factor

Factorising reverses expansion. Remove any common factor first; then, for $x^{2} + b x + c$ , find two numbers that multiply to $c$ and add to $b$ . Both conditions must hold.

If $c$ is negative, the numbers have opposite signs. If $c$ is positive, their signs match, and the sign of $b$ tells you whether both are positive or negative.

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

When the leading coefficient is not one

For $a x^{2} + b x + c$ , split the middle term using numbers $p$ and $q$ with $p q = a c$ and $p + q = b$ . This preserves $b x$ while creating two groups that can share a bracket.

Factor each group, then factor out the matching bracket. If no integer pair works, an integer factorisation may not exist. Always expand your proposed factors to check all three coefficients.

\begin{array}{c} a x^{2} + b x + c = a x^{2} + p x + q x + c \\ p + q = b, p q = a c \end{array}

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Find a product and sum

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

Worked solution

1
Match the product and sum.
$(- 4) (3) = - 12, - 4 + 3 = - 1$
The constant term fixes the product; the linear coefficient fixes the sum.
2
Form the two factors.
$x^{2} - x - 12 = (x - 4) (x + 3)$
These signed constants supply the required cross-products and constant product.
3
Expand to verify.
$\begin{aligned} (x - 4) (x + 3) & = x^{2} + 3 x - 4 x - 12 \\ = x^{2} - x - 12 \end{aligned}$
Matching every coefficient confirms the factorisation.

Answer

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

EXAMPLE 02

Split a non-monic middle term

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

Worked solution

1
Find the split numbers.
$\begin{array}{c} a c = 6 (- 3) = - 18 \\ 9 (- 2) = - 18, 9 + (- 2) = 7 \end{array}$
The split numbers must multiply to $a c$ , rather than just to $c$ .
2
Split and group the middle term.
$\begin{aligned} 6 x^{2} + 7 x - 3 & = 6 x^{2} + 9 x - 2 x - 3 \\ = 3 x (2 x + 3) - 1 (2 x + 3) \end{aligned}$
Taking out a negative factor from the second group produces matching brackets.
3
Factor out the common bracket.
$6 x^{2} + 7 x - 3 = (3 x - 1) (2 x + 3)$
The entire expression $2 x + 3$ is common to both groups.
4
Check the expansion.
$\begin{aligned} (3 x - 1) (2 x + 3) & = 6 x^{2} + 9 x - 2 x - 3 \\ = 6 x^{2} + 7 x - 3 \end{aligned}$
The leading, linear and constant terms all agree with the original expression.

Answer

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

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

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

The numbers $3$ and $4$ have product $12$ and sum $7$ , so $x^{2} + 7 x + 12 = (x + 3) (x + 4)$ . Expanding gives the original three terms.

Keep practising

6 worksheet questions · 16 marks

Open worksheet 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 ↗

Monic and non-monic factorisation

In this lesson

Understand the idea.

Start with the simplest factor

When the leading coefficient is not one

See the method in action.

Find a product and sum

Match the product and sum.

Form the two factors.

Expand to verify.

Split a non-monic middle term

Find the split numbers.

Split and group the middle term.

Factor out the common bracket.

Check the expansion.

Try it yourself.

Check your understanding

Keep practising