Difference of Two Squares

The identity $(a + b)(a - b) = a^2 - b^2\text{:}$ why the middle terms cancel, what it looks like as an area, how to expand with it, and how it turns $103 \times 97$ or $2022^2 - 2021^2$ into easy sums.

Before you start

In lesson 3.5 you met the perfect squares $(a + b)^2 = a^2 + 2ab + b^2$ and $(a - b)^2 = a^2 - 2ab + b^2\text{.}$ This lesson multiplies the two different brackets, $(a + b)$ and $(a - b)$ , together.

The Identity

Expand $(a + b)(a - b)$ , each term by each term:

$(a + b)(a - b) = a^2 - ab + ba - b^2 = a^2 - ab + ab - b^2 = a^2 - b^2$

The two middle terms, $-ab$ and $+ab$ , are opposites, so they cancel. In the frame they lie on one diagonal:

×	$a$	$-b$
$a$	$a^2$	$-ab$
$+b$	$+ab$	$-b^2$

$a^2 - ab + ab - b^2 = a^2 - b^2$

The difference of two squares

$(a + b)(a - b) = a^2 - b^2$

The sum of two terms times their difference is the square of the first term minus the square of the second. The order of the brackets does not matter: $(a - b)(a + b) = a^2 - b^2$ too.

It is an identity: true for all values of $a$ and $b$ , and for any terms in their place.

Seeing It as an Area

Take $a$ greater than $b$ . (i) Start with a square of side $a$ and mark a square of side $b$ in one corner. (ii) Remove the small square: the L-shape that is left has area $a^2 - b^2$ . Cut it along the dashed line into piece 1, $a$ long and $(a - b)$ high, and piece 2, $(a - b)$ long and $b$ high. (iii) Turn piece 2 on its side and put it at the end of piece 1. Together they make a rectangle $(a + b)$ long and $(a - b)$ high.

Nothing was added or lost in moving the pieces, so the areas are equal: $a^2 - b^2 = (a + b)(a - b)\text{.}$

Expanding with the Identity

Check that the two brackets have the same two terms, one with $+$ and one with $-$ between them. Then the answer is (first term) $^2$ minus (second term) $^2$ , with no middle term.

Example 1: One letter, either order

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

Solution

(a) First term $x$ , second term $5$ : $x^2 - 5^2 = x^2 - 25$ .

In full: $x^2 - 5x + 5x - 25 = x^2 - 25\text{.}$ The identity skips the cancelling terms.

(b) The minus bracket is written first, which does not matter: $(5y)^2 - 4^2 = 25y^2 - 16\text{.}$

(a) $x^2 - 25$ ; (b) $25y^2 - 16$

Example 2: Coefficients and two letters

Expand (a) $(2x + 7y)(2x - 7y)$ ; (b) $(6b - a)(6b + a)$ .

Solution

(a) The first term is $2x$ and the second is $7y$ : $(2x)^2 - (7y)^2 = 4x^2 - 49y^2\text{.}$

(b) The first term is $6b$ and the second is $a$ : $(6b)^2 - a^2 = 36b^2 - a^2\text{.}$

(a) $4x^2 - 49y^2$ ; (b) $36b^2 - a^2$

Example 3: A fraction

Expand $(4a - \frac{7}{2}b)(4a + \frac{7}{2}b)\text{.}$

Solution

The first term is $4a$ and the second is $\frac{7}{2}b$ .

$(4a)^2 - (\frac{7}{2}b)^2 = 16a^2 - \frac{49}{4}b^2$

$16a^2 - \frac{49}{4}b^2$

Example 4: Terms with powers, and the number first

Expand (a) $(x^2 + 3)(x^2 - 3)$ ; (b) $(ab + c)(ab - c)$ ; (c) $(5 + x)(5 - x)$ .

Solution

(a) The first term is $x^2$ : $(x^2)^2 - 3^2 = x^4 - 9$ , since $x^2 \times x^2 = x \times x \times x \times x = x^4\text{.}$

(b) The first term is $ab$ : $(ab)^2 - c^2 = a^2b^2 - c^2\text{.}$

(a) $x^4 - 9$ ; (b) $a^2b^2 - c^2$ ; (c) $25 - x^2$

Example 5: When it is not a difference of two squares

Expand $(x - 5)(5 - x)$ .

Solution

The brackets have the same numbers, but they are not the sum and the difference of the same two terms: $5 - x = -(x - 5)$ . So the identity does not apply.

Expand in full: $5x - x^2 - 25 + 5x = -x^2 + 10x - 25\text{.}$

This is $-(x - 5)^2$ : the product of $(x - 5)$ and its negative.

$(x - 5)(5 - x) = -x^2 + 10x - 25$

Example 6: With a perfect square

Simplify $(x + 3)^2 - (x + 3)(x - 3)\text{.}$

Solution

$(x + 3)^2 = x^2 + 6x + 9$ and $(x + 3)(x - 3) = x^2 - 9\text{.}$

$(x^2 + 6x + 9) - (x^2 - 9) = x^2 + 6x + 9 - x^2 + 9 = 6x + 18$

$6x + 18$

Example 7: Spot the error

Ravi wrote $(x + 4)(x - 4) = x^2 - 8x - 16\text{.}$ Check with $x = 1$ , then correct it.

Solution

$(5)(-3) = -15$ , but $1 - 8 - 16 = -23$ . Wrong.

The middle terms $-4x$ and $+4x$ cancel; there is no $-8x$ . Correct: $x^2 - 16$ , and $1 - 16 = -15$ . ✓

$(x + 4)(x - 4) = x^2 - 16$

Number Work

Two numbers the same distance either side of a round number can be multiplied with the identity. And because the identity works both ways, $a^2 - b^2$ can be worked out as $(a + b)(a - b)$ .

Example 8: Products near a round number

Without a calculator, find (a) $103 \times 97$ ; (b) $51 \times 49$ .

Solution

(a) $103 \times 97 = (100 + 3)(100 - 3) = 100^2 - 3^2 = 10\,000 - 9 = 9991$

(b) $51 \times 49 = (50 + 1)(50 - 1) = 2500 - 1 = 2499$

(a) $9991$ ; (b) $2499$

Example 9: Differences of squares of numbers

Without a calculator, find (a) $65^2 - 35^2$ ; (b) $2022^2 - 2021^2$ .

Solution

(a) $65^2 - 35^2 = (65 + 35)(65 - 35) = 100 \times 30 = 3000$

(b) $2022^2 - 2021^2 = (2022 + 2021)(2022 - 2021) = 4043 \times 1 = 4043$

(a) $3000$ ; (b) $4043$

Consecutive squares

In the same way, $n^2 - (n - 1)^2 = (n + n - 1)(n - n + 1) = 2n - 1\text{.}$ So two consecutive squares always differ by an odd number: $10^2 - 9^2 = 19$ , $50^2 - 49^2 = 99$ , $100^2 - 99^2 = 199$ .

Common Mistakes

Writing a middle term: $(x + 4)(x - 4) = x^2 - 16\text{,}$ not $x^2 - 8x - 16$ .
Squaring only the letter: $(3x + 2)(3x - 2) = 9x^2 - 4\text{,}$ not $3x^2 - 4$ .
A plus sign in the answer: $(y - 6)(y + 6) = y^2 - 36\text{,}$ not $y^2 + 36$ .
Using the identity when the brackets are not a sum and difference of the same terms: $(x - 5)(5 - x)$ is $-x^2 + 10x - 25$ .
Confusing $(a + b)(a - b) = a^2 - b^2$ with $(a - b)^2 = a^2 - 2ab + b^2\text{.}$

Practice

Try each question before you open its solution.

Question 1

Expand (a) $(x + 3)(x - 3)$ ; (b) $(y - 8)(y + 8)$ ; (c) $(2p + 1)(2p - 1)$ .

ShowHide solution

(a) $x^2 - 3^2 = x^2 - 9$

(b) $y^2 - 8^2 = y^2 - 64$

(a) $x^2 - 9$ ; (b) $y^2 - 64$ ; (c) $4p^2 - 1$

Question 2

Expand (a) $(3a + 5b)(3a - 5b)$ ; (b) $(7m - n)(7m + n)$ .

ShowHide solution

(a) $(3a)^2 - (5b)^2 = 9a^2 - 25b^2$

(b) $(7m)^2 - n^2 = 49m^2 - n^2$

(a) $9a^2 - 25b^2$ ; (b) $49m^2 - n^2$

Question 3

Expand (a) $(x + \frac{1}{2})(x - \frac{1}{2})\text{;}$ (b) $(\frac{2}{3}y - 3)(\frac{2}{3}y + 3)\text{.}$

ShowHide solution

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

(b) $(\frac{2}{3}y)^2 - 3^2 = \frac{4}{9}y^2 - 9$

(a) $x^2 - \frac{1}{4}$ ; (b) $\frac{4}{9}y^2 - 9$

Question 4

Expand (a) $(y^2 + 5)(y^2 - 5)$ ; (b) $(2xy - 1)(2xy + 1)$ .

ShowHide solution

(a) $(y^2)^2 - 5^2 = y^4 - 25$

(b) $(2xy)^2 - 1^2 = 4x^2y^2 - 1$

(a) $y^4 - 25$ ; (b) $4x^2y^2 - 1$

Question 5

Expand (a) $(6 - x)(6 + x)$ ; (b) $(1 + 3k)(1 - 3k)$ .

ShowHide solution

(a) $6^2 - x^2 = 36 - x^2$

(b) $1^2 - (3k)^2 = 1 - 9k^2$

(a) $36 - x^2$ ; (b) $1 - 9k^2$

Question 6

Which of these products is a difference of two squares? Expand both. (a) $(x + 2)(2 - x)$ ; (b) $(x - 3)(3 - x)$

ShowHide solution

(a) $(x + 2)(2 - x) = (2 + x)(2 - x)\text{:}$ the sum and the difference of $2$ and $x$ . So it is $4 - x^2$ .

(b) $3 - x = -(x - 3)$ , so this is not a sum times a difference. In full: $3x - x^2 - 9 + 3x = -x^2 + 6x - 9\text{.}$

(a) is a difference of two squares: $4 - x^2$ . (b) is not: $-x^2 + 6x - 9$ .

Question 7

Simplify (a) $(x + 4)^2 - (x + 4)(x - 4)\text{;}$ (b) $(2x - 1)(2x + 1) - (2x - 1)^2\text{.}$

ShowHide solution

(a) $(x^2 + 8x + 16) - (x^2 - 16) = 8x + 32$

(b) $(4x^2 - 1) - (4x^2 - 4x + 1) = 4x^2 - 1 - 4x^2 + 4x - 1 = 4x - 2$

(a) $8x + 32$ ; (b) $4x - 2$

Question 8

Without a calculator, find (a) $102 \times 98$ ; (b) $45 \times 35$ ; (c) $9.8 \times 10.2$ .

ShowHide solution

(a) $(100 + 2)(100 - 2) = 10\,000 - 4 = 9996$

(b) $(40 + 5)(40 - 5) = 1600 - 25 = 1575$

(a) $9996$ ; (b) $1575$ ; (c) $99.96$

Question 9

Without a calculator, find (a) $75^2 - 25^2$ ; (b) $1001^2 - 999^2$ ; (c) $40^2 - 39^2$ .

ShowHide solution

(a) $(75 + 25)(75 - 25) = 100 \times 50 = 5000$

(b) $(1001 + 999)(1001 - 999) = 2000 \times 2 = 4000$

(a) $5000$ ; (b) $4000$ ; (c) $79$

Question 10

A square card has sides of $x$ cm. A square of side $3$ cm is cut from one corner. (a) Write the area left as a difference of two squares. (b) Show that it equals $(x + 3)(x - 3)$ . (c) Find the area left when $x = 13$ , in two ways.

ShowHide solution

(a) $x^2 - 3^2 = x^2 - 9$ cm².

(b) $(x + 3)(x - 3) = x^2 - 3x + 3x - 9 = x^2 - 9\text{.}$ ✓

(a) $(x^2 - 9)$ cm²; (c) $160$ cm²

Question 11

A square plot has sides of $(x + 2)$ m. A rectangular plot is $(x + 4)$ m long and $x$ m wide. Show that the square plot is always $4$ m² larger.

ShowHide solution

The rectangle’s sides are $2$ m more and $2$ m less than $(x + 2)$ : $x + 4 = (x + 2) + 2$ and $x = (x + 2) - 2$ .

So its area is $((x + 2) + 2)((x + 2) - 2) = (x + 2)^2 - 4\text{,}$ by the difference of two squares.

The square’s area is $(x + 2)^2$ , which is $4$ m² more. Check with $x = 3$ : $5^2 = 25$ and $7 \times 3 = 21$ ; $25 - 21 = 4$ . ✓

The square plot is $4$ m² larger for every value of $x$ .

Question 12

Each line has one mistake. Find it and correct it. (a) $(5x + 3)(5x - 3) = 5x^2 - 9\text{;}$ (b) $(k - 7)(k + 7) = k^2 + 49\text{;}$ (c) $(a + b)(a - b) = a^2 - 2ab - b^2$

ShowHide solution

(a) The first term must be squared whole: $(5x)^2 = 25x^2$ . Correct: $25x^2 - 9$ .

(b) The last term is $(+7) \times (-7) = -49\text{.}$ Correct: $k^2 - 49$ .

(c) The middle terms $-ab$ and $+ab$ cancel. Correct: $a^2 - b^2$ . Check with $a = 3$ , $b = 1$ : $(4)(2) = 8 = 9 - 1$ . ✓

(a) $25x^2 - 9$ ; (b) $k^2 - 49$ ; (c) $a^2 - b^2$

Key Points

$(a + b)(a - b) = a^2 - b^2\text{,}$ in either order of brackets; the middle terms cancel.
As an area: the L-shape $a^2 - b^2$ cut and moved makes a rectangle $(a + b)$ by $(a - b)$ .
Square the whole of each term: $(2x + 7y)(2x - 7y) = 4x^2 - 49y^2\text{.}$
The brackets must be the sum and the difference of the same two terms; $(x - 5)(5 - x)$ is not.
Number work: $103 \times 97 = 100^2 - 3^2\text{,}$ and $65^2 - 35^2 = 100 \times 30\text{.}$
$n^2$ and $(n - 1)^2$ differ by $2n - 1$ , an odd number.