YEAR 10 · QUADRATICS AND PARABOLAS · 14 MIN

Expanding quadratic products

Expand two linear brackets accurately and recognise square and difference-of-squares patterns.

In this lesson

Multiply every pair of terms in a bracket product.
Recognise useful quadratic product patterns.

Before you start

Combining like terms
Multiplying algebraic terms

01 / LEARN

Understand the idea.

Multiply all four pairs

Expanding uses the distributive law twice: each term in the first bracket multiplies both terms in the second. Treat a negative term, such as $- 3$ , as one signed quantity.

Write all four products before collecting like terms. Only terms with the same variable part combine: $x^{2}$ and $x$ remain separate. The product table records the same four multiplications.

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

Recognise a pattern, then check it

A squared bracket means two identical brackets multiplied together. Its two cross-products are equal, which explains the middle term in ${(x + p)}^{2} = x^{2} + 2 p x + p^{2}$ .

In $(x + p) (x - p)$ , the cross-products are opposites and cancel. The constant product is $- p^{2}$ , so this pattern is a difference of squares. Direct expansion checks either shortcut.

\begin{aligned} {(x + p)}^{2} & = x^{2} + 2 p x + p^{2} \\ (x + p) (x - p) & = x^{2} - p^{2} \end{aligned}

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

One positive and one negative constant

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

Each cell records one product.
×	x	−3
x	x²	−3x
4	4x	−12

Worked solution

1
Distribute the first term.
$x (x - 3) = x^{2} - 3 x$
The first term multiplies both terms in the second bracket.
2
Distribute the constant.
$4 (x - 3) = 4 x - 12$
A positive number times a negative number gives a negative product.
3
Collect the like terms.
$\begin{aligned} (x + 4) (x - 3) & = x^{2} - 3 x + 4 x - 12 \\ = x^{2} + x - 12 \end{aligned}$
The linear terms combine because $- 3 + 4 = 1$ .

Answer

$x^{2} + x - 12$

EXAMPLE 02

Non-monic brackets

Expand $(2 x - 5) (3 x + 2)$ .

Worked solution

1
Multiply by the first term.
$2 x (3 x + 2) = 6 x^{2} + 4 x$
Multiplying $x$ by $x$ gives $x^{2}$ .
2
Multiply by the signed constant.
$- 5 (3 x + 2) = - 15 x - 10$
Both products retain the negative sign from the first bracket.
3
Combine the linear terms.
$\begin{aligned} (2 x - 5) (3 x + 2) & = 6 x^{2} + 4 x - 15 x - 10 \\ = 6 x^{2} - 11 x - 10 \end{aligned}$
The coefficient of $x$ is $4 - 15 = - 11$ .

Answer

$6 x^{2} - 11 x - 10$

03 / YOUR TURN

Try it yourself.

Check your understanding

Expand $(x + 2) (x + 7)$ .

Show answer & reasoning

$x^{2} + 9 x + 14$

The four products are $x^{2}$ , $7 x$ , $2 x$ and $14$ . Combining the two linear terms gives $x^{2} + 9 x + 14$ .

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

Expand $(x + 3) (x + 5)$ .
2 marks
Expand $(x - 6) (x + 2)$ .
2 marks
Expand $(2 x + 1) (x - 4)$ .
2 marks
Expand ${(3 x - 2)}^{2}$ .
2 marks
A rectangle has side lengths $(x + 2)$ cm and $(x + 5)$ cm. Write its area as an expanded expression.
3 marks
A student says $(x - 4) (x + 4) = x^{2} + 16$ . Explain the sign error and give the correct expansion.
3 marks

Read the worked solutions

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

Expand $(x + 3) (x + 5)$ .
2 marks
$x^{2} + 5 x + 3 x + 15 = x^{2} + 8 x + 15$ .
$x^{2} + 8 x + 15$
Expand $(x - 6) (x + 2)$ .
2 marks
$x^{2} + 2 x - 6 x - 12 = x^{2} - 4 x - 12$ .
$x^{2} - 4 x - 12$
Expand $(2 x + 1) (x - 4)$ .
2 marks
$2 x^{2} - 8 x + x - 4 = 2 x^{2} - 7 x - 4$ .
$2 x^{2} - 7 x - 4$
Expand ${(3 x - 2)}^{2}$ .
2 marks
Use ${(3 x)}^{2} - 2 (3 x) (2) + 2^{2}$ .
$9 x^{2} - 12 x + 4$
A rectangle has side lengths $(x + 2)$ cm and $(x + 5)$ cm. Write its area as an expanded expression.
3 marks
Area is $(x + 2) (x + 5) = x^{2} + 7 x + 10$ .
The unit is square centimetres.
$(x^{2} + 7 x + 10) {cm}^{2}$
A student says $(x - 4) (x + 4) = x^{2} + 16$ . Explain the sign error and give the correct expansion.
3 marks
The middle terms $4 x$ and $- 4 x$ cancel.
The last product is $(- 4) (4) = - 16$ .
$x^{2} - 16$ ; the constant product is negative.

Checking saved progress…

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 ↗

In this lesson

Understand the idea.

Multiply all four pairs

Recognise a pattern, then check it

See the method in action.

One positive and one negative constant

Distribute the first term.

Distribute the constant.

Collect the like terms.

Non-monic brackets

Multiply by the first term.

Multiply by the signed constant.

Combine the linear terms.

Try it yourself.

Check your understanding

Keep practising