Expanding two brackets | Year 9

01 / LEARN

The idea, step by step

Every term has a partner

In $(x + 3) (x + 5)$ , the first bracket means an entire quantity is multiplied by another entire quantity. Distribute each term in the first bracket to both terms in the second. A four-cell product grid is a useful way to keep track: $x \times x$ , $x \times 5$ , $3 \times x$ and $3 \times 5$ .

The two cross-products, $5 x$ and $3 x$ , are like terms. Add their coefficients, but keep the $x$ . The result is $x^{2} + 8 x + 15$ . A quadratic expression has highest power $2$ in its variable, with a non-zero squared-term coefficient; it need not have a visible constant or linear term.

Keep the signs attached

Treat subtraction as adding a negative term. For $(x - 2) (x + 7)$ , the products are $x^{2}$ , $7 x$ , $- 2 x$ and $- 14$ . The middle terms combine to $5 x$ . Do not change the sign of the final product just because the first term is positive.

Check a proposed expansion by substituting an easy value such as $x = 1$ into both forms. This can reveal an error, although checking one value alone does not prove that two expressions are always equal.

A common trap

$(x + 3) (x + 5) = x^{2} + 15$ .

Instead: The cross-products also count: $5 x + 3 x = 8 x$ .

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example

Expand $(x + 4) (x + 2)$ .

Distribute: $x (x + 2) + 4 (x + 2)$ .
Write all four products: $x^{2} + 2 x + 4 x + 8$ .
Collect the like terms: $2 x + 4 x = 6 x$ .

x^{2} + 6 x + 8

Example 2

Worked example

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

The products are $2 x^{2}$ , $10 x$ , $- 3 x$ and $- 15$ .
Combine the middle terms: $10 x - 3 x = 7 x$ .
At $x = 0$ , both expressions give $- 15$ , which checks the constant term.

2 x^{2} + 7 x - 15

Check your understanding

Expand $(x + 1) (x + 6)$ .

Show answer

$x^{2} + 7 x + 6$

The four products are $x^{2}$ , $6 x$ , $x$ and $6$ .

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Expand $(x + 2) (x + 3)$ .
2 marks
Expand $(x - 4) (x + 1)$ .
2 marks
Expand $(x - 5) (x - 2)$ .
2 marks
Expand $(3 x + 2) (x - 1)$ .
2 marks
A rectangle has sides $(x + 2)$ cm and $(x + 7)$ cm. Write its area as an expanded expression.
3 marks
Sam says $(x + 2) (x - 2) = x^{2} - 4 x - 4$ . Correct the expansion and explain the middle term.
3 marks

Read the worked solutions

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

Expand $(x + 2) (x + 3)$ .
2 marks
$x^{2} + 3 x + 2 x + 6 = x^{2} + 5 x + 6$ .
$x^{2} + 5 x + 6$
Expand $(x - 4) (x + 1)$ .
2 marks
$x^{2} + x - 4 x - 4 = x^{2} - 3 x - 4$ .
$x^{2} - 3 x - 4$
Expand $(x - 5) (x - 2)$ .
2 marks
$x^{2} - 2 x - 5 x + 10 = x^{2} - 7 x + 10$ .
$x^{2} - 7 x + 10$
Expand $(3 x + 2) (x - 1)$ .
2 marks
$3 x^{2} - 3 x + 2 x - 2 = 3 x^{2} - x - 2$ .
$3 x^{2} - x - 2$
A rectangle has sides $(x + 2)$ cm and $(x + 7)$ cm. Write its area as an expanded expression.
3 marks
$A = (x + 2) (x + 7) = x^{2} + 7 x + 2 x + 14$ .
$(x^{2} + 9 x + 14) {cm}^{2}$
Sam says $(x + 2) (x - 2) = x^{2} - 4 x - 4$ . Correct the expansion and explain the middle term.
3 marks
$x^{2} - 2 x + 2 x - 4 = x^{2} - 4$ .
The cross-products cancel because $- 2 x + 2 x = 0$ .
$x^{2} - 4$

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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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