Expanding and factorising quadratics | Year 11

01 / LEARN

Understand the idea.

Expand systematically

Expanding uses the distributive law: every term in the first bracket multiplies every term in the second. Write all products with their signs, then collect terms with matching powers. Only like terms can be combined.

A squared bracket is a product of two identical brackets. Thus ${(x + a)}^{2} = x^{2} + 2 a x + a^{2}$ , including the middle term. In $(x - a) (x + a)$ , the middle terms cancel, giving the difference of squares $x^{2} - a^{2}$ .

(x + a) (x + b) = x^{2} + (a + b) x + a b

Reverse a product

Factorising reverses expansion without changing the expression’s value. Remove any common factor first. For a monic quadratic $x^{2} + b x + c$ , find two numbers with sum $b$ and product $c$ , then use them as the constants in the brackets.

For $a x^{2} + b x + c$ , split the middle term using two coefficients whose sum is $b$ and product is $a c$ . Group the four terms to expose a shared bracket. Expanding your final factors checks both signs and coefficients.

a x^{2} + b x + c = (p x + q) (r x + s)

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A negative constant

Expand and simplify $(x - 4) (x + 7)$ .

Worked solution

1
Multiply the first term by both terms in the second bracket.
$x (x + 7) = x^{2} + 7 x$
The distributive law includes both products involving the first x.
2
Multiply the remaining term by both terms.
$- 4 (x + 7) = - 4 x - 28$
The negative sign belongs to the entire term negative four.
3
Combine the products and collect like terms.
$\begin{aligned} (x - 4) (x + 7) & = x^{2} + 7 x - 4 x - 28 \\ = x^{2} + 3 x - 28 \end{aligned}$
Only the two terms proportional to x combine; the quadratic and constant terms retain their powers.

Answer

$x^{2} + 3 x - 28$

EXAMPLE 02

A non-monic quadratic

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

Worked solution

1
Find coefficients for splitting the middle term.
$a c = 6 (- 2) = - 12, 4 + (- 3) = 1, 4 (- 3) = - 12$
The split must reproduce the middle coefficient and enable grouping.
2
Split and group the expression.
$\begin{aligned} 6 x^{2} + x - 2 & = 6 x^{2} + 4 x - 3 x - 2 \\ = 2 x (3 x + 2) - (3 x + 2) \end{aligned}$
The first pair has common factor two x and the second pair has common factor negative one.
3
Factor out the shared bracket.
$6 x^{2} + x - 2 = (2 x - 1) (3 x + 2)$
The common factor is the complete bracket three x plus two.
4
Check by expanding.
$(2 x - 1) (3 x + 2) = 6 x^{2} + 4 x - 3 x - 2 = 6 x^{2} + x - 2$
Expansion recovers the original quadratic, including its middle coefficient.

Answer

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

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

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

The pair $- 4$ and $3$ has product $- 12$ and sum $- 1$ . Therefore $x^{2} - x - 12 = (x - 4) (x + 3)$ ; expansion checks the middle term.

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

Expand $(x + 3) (x + 8)$ .
2 marks
Expand $(2 x - 5) (x + 4)$ .
2 marks
Factorise $x^{2} + 2 x - 35$ .
2 marks
Factorise $4 x^{2} - 25$ .
2 marks
Factorise $3 x^{2} - 12 x$ .
2 marks
Factorise $2 x^{2} - 7 x + 3$ .
3 marks

Read the worked solutions

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

Expand $(x + 3) (x + 8)$ .
2 marks
The products are $x^{2} + 8 x + 3 x + 24$ .
Collect the linear terms.
$x^{2} + 11 x + 24$
Expand $(2 x - 5) (x + 4)$ .
2 marks
Multiply: $2 x^{2} + 8 x - 5 x - 20$ .
Collect $8 x - 5 x$ .
$2 x^{2} + 3 x - 20$
Factorise $x^{2} + 2 x - 35$ .
2 marks
Find product -35 and sum 2: 7 and -5.
Write the corresponding brackets.
$(x + 7) (x - 5)$
Factorise $4 x^{2} - 25$ .
2 marks
Recognise ${(2 x)}^{2} - 5^{2}$ .
Use difference of squares.
$(2 x - 5) (2 x + 5)$
Factorise $3 x^{2} - 12 x$ .
2 marks
The common factor is $3 x$ .
Divide both terms by it.
$3 x (x - 4)$
Factorise $2 x^{2} - 7 x + 3$ .
3 marks
Split $- 7 x$ into $- 6 x - x$ .
Group $2 x (x - 3) - 1 (x - 3)$ .
$(2 x - 1) (x - 3)$

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics topics 2.1–2.2: quadratic relationships, cubic and quartic polynomials. Original Ascent teaching sequence.

Official curriculum source ↗

In this lesson

Understand the idea.

Expand systematically

Reverse a product

See the method in action.

A negative constant

Multiply the first term by both terms in the second bracket.

Multiply the remaining term by both terms.

Combine the products and collect like terms.

A non-monic quadratic

Find coefficients for splitting the middle term.

Split and group the expression.

Factor out the shared bracket.

Check by expanding.

Try it yourself.

Check your understanding

Keep practising