Expanding Two Brackets

How to expand a product such as $(x + 2)(x + 3)$ or $(3 - 2x)(3x + 1)$ : each term in the first bracket times each term in the second, seen as the area of a rectangle and set out in a multiplication frame.

Before you start

In the last lesson you expanded a term times a bracket: $x(x + 3) = x^2 + 3x$ and $-2(x - 4) = -2x + 8\text{.}$ The term outside multiplies every term inside.

Signs: like signs multiply to give $+$ , unlike signs give $-$ . So $(-3) \times (-5) = 15$ and $(-3) \times 5 = -15$ .

Why There Are Four Terms

Think of $(a + b)(c + d)$ as the area of a rectangle with sides $(a + b)$ and $(c + d)$ . Cutting the sides at the $+$ signs cuts the rectangle into four pieces, one for each term of the first bracket with each term of the second.

Expanding two brackets

$(a + b)(c + d) = ac + ad + bc + bd$

Multiply each term in the first bracket by each term in the second, then collect like terms.

The same comes from the distributive law used twice: $(a + b)(c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd\text{.}$

The rectangle needs $a$ , $b$ , $c$ and $d$ to be positive lengths, but the result is true for all values, including negative ones.

The Multiplication Frame

The multiplication frame (the grid method) sets out the same four products in a table. Write the terms of the first bracket down the left and the terms of the second along the top, each with its sign. Each cell is its row times its column. Then add the four cells.

Example 1: Expand $(x + 2)(x + 3)$

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

Solution

As a rectangle, the four pieces are the four products:

In a frame: $x + 2$ down the left, $x + 3$ along the top.

×	$x$	$+3$
$x$	$x^2$	$+3x$
$+2$	$+2x$	$+6$

$x^2 + 3x + 2x + 6 = x^2 + 5x + 6$

Row $x$ : $x \times x = x^2$ and $x \times 3 = 3x$ . Row $+2$ : $2 \times x = 2x$ and $2 \times 3 = 6$ .

The two $x$ terms, $3x$ and $2x$ , lie on one diagonal of the frame and are like terms: $3x + 2x = 5x$ .

$(x + 2)(x + 3) = x^2 + 5x + 6$

Negative Terms

A term keeps its sign when it goes into the frame: $(x - 2)$ gives the terms $x$ and $-2$ . Each cell then follows the sign rules.

Example 2: The four sign patterns

Expand $(x + 4)(x - 2)$ with a frame. Then compare it with $(x + 4)(x + 2)$ , $(x - 4)(x + 2)$ and $(x - 4)(x - 2)$ .

Solution

×	$x$	$-2$
$x$	$x^2$	$-2x$
$+4$	$+4x$	$-8$

$x^2 - 2x + 4x - 8 = x^2 + 2x - 8$

The other three in the same way:

$(x + 4)(x + 2) = x^2 + 2x + 4x + 8 = x^2 + 6x + 8$

$(x - 4)(x + 2) = x^2 + 2x - 4x - 8 = x^2 - 2x - 8$

$(x - 4)(x - 2) = x^2 - 2x - 4x + 8 = x^2 - 6x + 8$

In each, the number term is the product of the two numbers, and the coefficient of $x$ is their sum: for $(x - 4)(x + 2)$ , $(-4) \times 2 = -8$ and $-4 + 2 = -2$ .

$(x + 4)(x - 2) = x^2 + 2x - 8\text{.}$ The others are $x^2 + 6x + 8$ , $x^2 - 2x - 8$ and $x^2 - 6x + 8$ .

Example 3: A coefficient in front of $x$

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

Solution

×	$x$	$-2$
$3x$	$3x^2$	$-6x$
$+1$	$+x$	$-2$

$3x^2 - 6x + x - 2 = 3x^2 - 5x - 2$

$3x \times x = 3x^2$ , $3x \times (-2) = -6x$ , $1 \times x = x$ , $1 \times (-2) = -2$ . Then $-6x + x = -5x$ .

$(3x + 1)(x - 2) = 3x^2 - 5x - 2$

Example 4: Without a frame

Expand $(3y - 4)(2y - 1)$ .

Solution

Each term of the first bracket times each term of the second, in order:

$3y \times 2y + 3y \times (-1) + (-4) \times 2y + (-4) \times (-1)$

$= 6y^2 - 3y - 8y + 4$

$= 6y^2 - 11y + 4$

$(3y - 4)(2y - 1) = 6y^2 - 11y + 4$

The Number First, or a Negative First Term

The method does not change when a bracket starts with a number or with a negative term. Only the answer needs tidying: write it in descending powers of the letter.

Example 5: Expand $(3 - 2x)(3x + 1)$

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

Solution

×	$3x$	$+1$
$3$	$9x$	$+3$
$-2x$	$-6x^2$	$-2x$

$9x + 3 - 6x^2 - 2x = -6x^2 + 7x + 3$

$3 \times 3x = 9x$ , $3 \times 1 = 3$ , $(-2x) \times 3x = -6x^2\text{,}$ $(-2x) \times 1 = -2x$ . The $x$ terms: $9x - 2x = 7x$ .

Check with $x = 1$ : $(3 - 2)(3 + 1) = 4$ , and $-6 + 7 + 3 = 4$ . ✓

$(3 - 2x)(3x + 1) = -6x^2 + 7x + 3$

Example 6: Both terms negative

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

Solution

$(-3x) \times 2x = -6x^2\text{,}$ $(-3x) \times 3 = -9x$ , $(-2) \times 2x = -4x$ , $(-2) \times 3 = -6$ .

$-6x^2 - 9x - 4x - 6 = -6x^2 - 13x - 6$

$(-3x - 2)(2x + 3) = -6x^2 - 13x - 6$

Fractions and Decimals

Example 7: Fraction and decimal coefficients

Expand (a) $(x + \frac{1}{2})(2x - 4)$ ; (b) $(0.5x + 1)(x - 2)$ .

Solution

(a) $x \times 2x = 2x^2$ , $x \times (-4) = -4x$ , $\frac{1}{2} \times 2x = x$ , $\frac{1}{2} \times (-4) = -2$ .

$2x^2 - 4x + x - 2 = 2x^2 - 3x - 2$

(b) $0.5x \times x = 0.5x^2$ , $0.5x \times (-2) = -x$ , $1 \times x = x$ , $1 \times (-2) = -2$ .

$0.5x^2 - x + x - 2 = 0.5x^2 - 2\text{.}$ The two $x$ terms cancel.

(a) $2x^2 - 3x - 2$ ; (b) $0.5x^2 - 2$

Checking by Substituting a Number

An expansion is equal to the product for every value of the letter. So put a number, such as $x = 2$ , into both. If the values differ, the expansion is wrong.

Example 8: Spot the error

Daniel wrote (a) $(x + 3)(x + 2) = x^2 + 6$ and (b) $(x - 4)(x + 1) = x^2 - 3x + 4\text{.}$ Check each with $x = 2$ , then correct it.

Solution

(a) Product: $(5)(4) = 20$ . Daniel’s answer: $4 + 6 = 10$ . Wrong: he multiplied only the first terms and the last terms, and left out $x \times 2$ and $3 \times x$ .

Correct: $x^2 + 2x + 3x + 6 = x^2 + 5x + 6\text{.}$ At $x = 2$ : $4 + 10 + 6 = 20$ . ✓

(b) Product: $(-2)(3) = -6$ . Daniel’s answer: $4 - 6 + 4 = 2$ . Wrong: $(-4) \times 1 = -4$ , not $+4$ .

Correct: $x^2 + x - 4x - 4 = x^2 - 3x - 4\text{.}$ At $x = 2$ : $4 - 6 - 4 = -6$ . ✓

(a) $x^2 + 5x + 6$ ; (b) $x^2 - 3x - 4$

Unknown Coefficients and Areas

Two expressions that are equal for every value of $x$ have the same coefficient of $x^2$ , the same coefficient of $x$ and the same number term. So we can compare them term by term.

Example 9: Find the unknown

Given that $(x + a)(x + 3) = x^2 + 9x + 18$ for all values of $x$ , find $a$ .

Solution

Expand the left side: $x^2 + 3x + ax + 3a = x^2 + (3 + a)x + 3a\text{.}$

Number terms: $3a = 18$ , so $a = 6$ .

Check the $x$ terms: $3 + 6 = 9$ . ✓

$a = 6$

Example 10: The area of a rectangle

A rectangle is $(x + 4)$ cm long and $(2x + 1)$ cm wide. Find an expression for its area, and its area when $x = 2$ .

Solution

Area $= (x + 4)(2x + 1) = 2x^2 + x + 8x + 4 = 2x^2 + 9x + 4$

When $x = 2$ : $2(4) + 18 + 4 = 30$ . Check: the rectangle is $6$ cm by $5$ cm, and $6 \times 5 = 30$ . ✓

The area is $(2x^2 + 9x + 4)$ cm², which is $30$ cm² when $x = 2$ .

Common Mistakes

Multiplying only the first terms and the last terms: $(x + 3)(x + 2) \ne x^2 + 6\text{.}$ There are four products.
Dropping a sign: in $(x - 4)(x + 1)$ the terms are $x$ and $-4$ , and $(-4) \times 1 = -4$ .
Adding the number terms instead of multiplying them: in $(3x + 2)(x + 5)$ the number term is $2 \times 5 = 10$ , not $7$ .
Forgetting to collect the two $x$ terms, or leaving the answer out of order: write $-6x^2 + 7x + 3$ , not $9x + 3 - 6x^2 - 2x$ .
Writing $(-2x) \times 3x$ as $6x^2$ . A negative times a positive is negative: $-6x^2$ .

Practice

Try each question before you open its solution. Give answers in descending powers.

Question 1

Expand $(x + 3)(x + 4)$ using a multiplication frame.

ShowHide solution

×	$x$	$+4$
$x$	$x^2$	$+4x$
$+3$	$+3x$	$+12$

$x^2 + 4x + 3x + 12 = x^2 + 7x + 12$

$x^2 + 7x + 12$

Question 2

Expand (a) $(x + 5)(x - 2)$ ; (b) $(x - 6)(x + 1)$ ; (c) $(x - 3)(x - 7)$ .

ShowHide solution

(a) $x^2 - 2x + 5x - 10 = x^2 + 3x - 10$

(b) $x^2 + x - 6x - 6 = x^2 - 5x - 6$

(a) $x^2 + 3x - 10$ ; (b) $x^2 - 5x - 6$ ; (c) $x^2 - 10x + 21$

Question 3

Expand (a) $(2x + 1)(x - 3)$ ; (b) $(4x - 3)(2x + 5)$ ; (c) $(2a + 1)(3a + 4)$ .

ShowHide solution

(a) $2x^2 - 6x + x - 3 = 2x^2 - 5x - 3$

(b) $8x^2 + 20x - 6x - 15 = 8x^2 + 14x - 15$

(a) $2x^2 - 5x - 3$ ; (b) $8x^2 + 14x - 15$ ; (c) $6a^2 + 11a + 4$

Question 4

Expand (a) $(3 - 2p)(4 - 3p)$ ; (b) $(4 - x)(3 + 2x)$ .

ShowHide solution

(a) $3 \times 4 = 12$ , $3 \times (-3p) = -9p$ , $(-2p) \times 4 = -8p$ , $(-2p) \times (-3p) = +6p^2\text{.}$

$12 - 9p - 8p + 6p^2 = 6p^2 - 17p + 12$

(b) $12 + 8x - 3x - 2x^2 = -2x^2 + 5x + 12$

(a) $6p^2 - 17p + 12$ ; (b) $-2x^2 + 5x + 12$

Question 5

Expand (a) $(-x + 4)(2x - 1)$ ; (b) $(-2y - 5)(y - 3)$ .

ShowHide solution

(a) $(-x)(2x) = -2x^2$ , $(-x)(-1) = +x$ , $4(2x) = 8x$ , $4(-1) = -4$ : $-2x^2 + 9x - 4$ .

(b) $(-2y)(y) = -2y^2$ , $(-2y)(-3) = +6y\text{,}$ $(-5)(y) = -5y$ , $(-5)(-3) = +15$ : $-2y^2 + y + 15$ .

(a) $-2x^2 + 9x - 4$ ; (b) $-2y^2 + y + 15$

Question 6

Expand (a) $(x - \frac{1}{2})(4x + 2)$ ; (b) $(0.2x + 3)(x + 5)$ .

ShowHide solution

(a) $4x^2 + 2x - 2x - 1 = 4x^2 - 1\text{,}$ since $-\frac{1}{2} \times 4x = -2x$ and $-\frac{1}{2} \times 2 = -1$ .

(b) $0.2x^2 + x + 3x + 15 = 0.2x^2 + 4x + 15\text{,}$ since $0.2x \times 5 = x$ .

(a) $4x^2 - 1$ ; (b) $0.2x^2 + 4x + 15$

Question 7

Given that $(2x + k)(x - 1) = 2x^2 + 3x - 5$ for all values of $x$ , find $k$ .

ShowHide solution

Expand: $2x^2 - 2x + kx - k = 2x^2 + (k - 2)x - k\text{.}$

Number terms: $-k = -5$ , so $k = 5$ .

Check the $x$ terms: $k - 2 = 3$ . ✓

$k = 5$

Question 8

Given that $(x + a)(x - 4) = x^2 + bx - 20$ for all values of $x$ , find $a$ and $b$ .

ShowHide solution

Expand: $x^2 - 4x + ax - 4a = x^2 + (a - 4)x - 4a\text{.}$

Number terms: $-4a = -20$ , so $a = 5$ . Then $b = a - 4 = 1$ .

Check: $(x + 5)(x - 4) = x^2 - 4x + 5x - 20 = x^2 + x - 20\text{.}$ ✓

$a = 5$ , $b = 1$

Question 9

The rectangle below is cut into four pieces. Write its area as the product of two brackets, then expand it.

ShowHide solution

The sides are $(x + 5)$ and $(x + 2)$ , so the area is $(x + 2)(x + 5)$ .

The four pieces are $x \times x = x^2$ , $x \times 5 = 5x$ , $2 \times x = 2x$ and $2 \times 5 = 10$ .

$x^2 + 5x + 2x + 10 = x^2 + 7x + 10$

$(x + 2)(x + 5) = x^2 + 7x + 10$

Question 10

A field is $(3x - 2)$ m long and $(x + 5)$ m wide. (a) Find an expression for its area. (b) Find the area when $x = 10$ .

ShowHide solution

(a) $(3x - 2)(x + 5) = 3x^2 + 15x - 2x - 10 = 3x^2 + 13x - 10$

(b) $3(100) + 130 - 10 = 420\text{.}$ Check: the field is $28$ m by $15$ m, and $28 \times 15 = 420$ . ✓

(a) $(3x^2 + 13x - 10)$ m²; (b) $420$ m²

Question 11

A square garden has sides of $x$ m. One pair of sides is made $3$ m longer and the other pair $1$ m shorter. (a) Find the new area. (b) By how much has the area increased? (c) Check your answers when $x = 6$ .

ShowHide solution

(a) $(x + 3)(x - 1) = x^2 - x + 3x - 3 = x^2 + 2x - 3$

(b) $(x^2 + 2x - 3) - x^2 = 2x - 3$

(c) When $x = 6$ the new garden is $9$ m by $5$ m, area $45$ m²; and $36 + 12 - 3 = 45$ . ✓ The old area was $36$ m², so the increase is $9$ m²; and $2(6) - 3 = 9$ . ✓

(a) $(x^2 + 2x - 3)$ m²; (b) $(2x - 3)$ m²; (c) $45$ m² and an increase of $9$ m²

Question 12

Each line has one mistake. Check each with $x = 2$ , then correct it. (a) $(x - 2)(x - 6) = x^2 - 8x - 12\text{;}$ (b) $(2x + 3)(x + 4) = 2x^2 + 11x + 7\text{;}$ (c) $(x + 1)(x + 1) = x^2 + 1$

ShowHide solution

(a) Product: $(0)(-4) = 0$ ; line: $4 - 16 - 12 = -24$ . $(-2) \times (-6) = +12\text{:}$ correct is $x^2 - 8x + 12$ , and $4 - 16 + 12 = 0$ . ✓

(b) Product: $(7)(6) = 42$ ; line: $8 + 22 + 7 = 37$ . The number term is $3 \times 4 = 12$ : correct is $2x^2 + 11x + 12$ , and $8 + 22 + 12 = 42$ . ✓

(c) Product: $(3)(3) = 9$ ; line: $4 + 1 = 5$ . The products $x \times 1$ and $1 \times x$ were left out: correct is $x^2 + 2x + 1$ , and $4 + 4 + 1 = 9$ . ✓

(a) $x^2 - 8x + 12$ ; (b) $2x^2 + 11x + 12$ ; (c) $x^2 + 2x + 1$

Key Points

$(a + b)(c + d) = ac + ad + bc + bd\text{:}$ each term in the first bracket times each term in the second.
The rectangle and the multiplication frame show the same four products; each cell is its row times its column.
Each term keeps its sign; like signs give $+$ , unlike signs give $-$ .
Collect the like terms and write the answer in descending powers.
For $(x + p)(x + q)$ , the number term is $pq$ and the coefficient of $x$ is $p + q$ .
Check an expansion by substituting a number into the product and the answer.
Equal expressions have equal coefficients, which lets you find an unknown such as $a$ in $(x + a)(x + 3)$ .

Expanding Two Brackets

Before you start

Why There Are Four Terms

Expanding two brackets

The Multiplication Frame

Example 1: Expand (x+2)(x+3)(x + 2)(x + 3)(x+2)(x+3)

Negative Terms

Example 2: The four sign patterns

Example 3: A coefficient in front of xxx

Example 4: Without a frame

The Number First, or a Negative First Term

Example 5: Expand (3−2x)(3x+1)(3 - 2x)(3x + 1)(3−2x)(3x+1)

Example 6: Both terms negative

Fractions and Decimals

Example 7: Fraction and decimal coefficients

Checking by Substituting a Number

Example 8: Spot the error

Unknown Coefficients and Areas

Example 9: Find the unknown

Example 10: The area of a rectangle

Common Mistakes

Practice

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Key Points

Example 1: Expand $(x + 2)(x + 3)$

Example 3: A coefficient in front of $x$

Example 5: Expand $(3 - 2x)(3x + 1)$