Factorising Using the Identities

How to recognise a difference of two squares and a perfect square, factorise them with the three identities, and tell when an identity does not fit.

Before you start

Factorising is expanding backwards: writing an expression as a product. In Secondary 1 you factorised by taking out a common factor: $6x + 9 = 3(2x + 3)$ .

In lessons 3.5 and 3.6 you expanded with three identities. Read from right to left, they factorise.

The Identities Read Backwards

Factorising with the identities

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

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

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

Always check a factorisation by expanding it back.

The skill is to recognise which identity an expression fits, and what $a$ and $b$ are. Two terms suggest a difference of two squares; three terms suggest a perfect square.

Recognising a Difference of Two Squares

A difference of two squares has

exactly two terms; each term a square (of a number, a letter, a term or a bracket); and a minus sign between them.

Find $a$ and $b$ by asking "what was squared?" Then write $(a + b)(a - b)$ .

Example 1: Squares of a letter and a number

Factorise (a) $x^2 - 9$ ; (b) $25x^2 - 16$ .

Solution

(a) $x^2$ is the square of $x$ and $9$ is the square of $3$ : $x^2 - 3^2 = (x + 3)(x - 3)\text{.}$

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

(b) $25x^2 = (5x)^2$ and $16 = 4^2$ : $(5x)^2 - 4^2 = (5x + 4)(5x - 4)\text{.}$

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

Example 2: The number first, and two letters

Factorise (a) $49 - 4y^2$ ; (b) $9x^2 - y^2$ ; (c) $a^2b^2 - 16$ .

Solution

(a) $49 = 7^2$ and $4y^2 = (2y)^2$ : $(7 + 2y)(7 - 2y)$ . Keep the order: the first square is $7^2$ .

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

(a) $(7 + 2y)(7 - 2y)$ ; (b) $(3x + y)(3x - y)$ ; (c) $(ab + 4)(ab - 4)$

Example 3: A bracket as one of the squares

Factorise (a) $(x + 1)^2 - 9$ ; (b) $4(a + b)^2 - 9$ .

Solution

(a) Here the first square is $(x + 1)^2$ , so $a = x + 1$ , and $b = 3$ .

$(x + 1)^2 - 3^2 = \big((x + 1) + 3\big)\big((x + 1) - 3\big) = (x + 4)(x - 2)$

Check: $(x + 4)(x - 2) = x^2 + 2x - 8\text{,}$ and $(x + 1)^2 - 9 = x^2 + 2x + 1 - 9 = x^2 + 2x - 8\text{.}$ ✓

(b) $4(a + b)^2 = \big(2(a + b)\big)^2$ and $9 = 3^2$ .

$\big(2(a + b) + 3\big)\big(2(a + b) - 3\big) = (2a + 2b + 3)(2a + 2b - 3)$

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

Example 4: A fraction

Factorise $x^2 - \frac{1}{4}$ .

Solution

$\frac{1}{4} = (\frac{1}{2})^2$ , so $x^2 - (\frac{1}{2})^2 = (x + \frac{1}{2})(x - \frac{1}{2})\text{.}$

$(x + \frac{1}{2})(x - \frac{1}{2})$

Recognising a Perfect Square

A perfect square has

three terms (put them in descending powers first);

a first term and a last term that are both squares, each with a $+$ sign in front;

a middle term equal to $2 \times$ (the square root of the first term) $\times$ (the square root of the last term), with a $+$ or $-$ sign.

The sign of the middle term is the sign inside the bracket: $a^2 + 2ab + b^2 = (a + b)^2$ and $a^2 - 2ab + b^2 = (a - b)^2\text{.}$

Example 5: Testing and factorising

Factorise (a) $x^2 + 10x + 25$ ; (b) $9y^2 + 24y + 16$ ; (c) $4a^2 - 12ab + 9b^2$ .

Solution

(a) $x^2 = x^2$ and $25 = 5^2$ . Middle: $2 \times x \times 5 = 10x$ . ✓ So $(x + 5)^2$ .

(b) $9y^2 = (3y)^2$ and $16 = 4^2$ . Middle: $2 \times 3y \times 4 = 24y$ . ✓ So $(3y + 4)^2$ .

(c) $4a^2 = (2a)^2$ and $9b^2 = (3b)^2$ . Middle: $2 \times 2a \times 3b = 12ab$ , and the sign is $-$ . So $(2a - 3b)^2$ .

Check (c): $(2a - 3b)^2 = 4a^2 - 12ab + 9b^2\text{.}$ ✓

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

Example 6: The number first

Factorise $16 - 8x + x^2$ .

Solution

$16 = 4^2$ and $x^2 = x^2$ , each with a $+$ sign. Middle: $2 \times 4 \times x = 8x$ , with a $-$ sign.

So $16 - 8x + x^2 = (4 - x)^2\text{.}$

In descending powers it is $x^2 - 8x + 16 = (x - 4)^2\text{.}$ Both answers are right: $(4 - x)^2 = (x - 4)^2$ , because a bracket and its negative have the same square.

$(4 - x)^2$ , which is the same as $(x - 4)^2$

Example 7: A fraction

Factorise $x^2 - x + \frac{1}{4}$ .

Solution

$x^2 = x^2$ and $\frac{1}{4} = (\frac{1}{2})^2$ . Middle: $2 \times x \times \frac{1}{2} = x$ , with a $-$ sign. ✓

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

When an Identity Does Not Fit

Example 8: Check before you factorise

Can these be factorised with an identity? Give a reason. (a) $x^2 + 9$ ; (b) $16x^2 + 28xy + 49y^2$ ; (c) $49h^2 - 42hk + 36k^2$

Solution

(a) Two squares, but with a plus sign between them. A sum of two squares such as $x^2 + 9$ cannot be factorised into two brackets: $(x + 3)(x - 3) = x^2 - 9$ and $(x + 3)^2 = x^2 + 6x + 9\text{.}$

(b) $16x^2 = (4x)^2$ and $49y^2 = (7y)^2$ , but $2 \times 4x \times 7y = 56xy$ , not $28xy$ . Not a perfect square.

None of them fits an identity: (a) is a sum of squares; in (b) and (c) the middle term is wrong for a perfect square.

A common factor can still come out

A sum of two squares such as $x^2 + 9$ or $4x^2 + 25y^2$ cannot be split into two brackets, but a common factor can still be taken out: $4x^2 + 4y^2 = 4(x^2 + y^2)\text{.}$

The Missing Term of a Perfect Square

Example 9: The missing term

Find the term that makes each a perfect square. (a) $x^2 + 6x + \square$ ; (b) $x^2 + \square x + 49$ ; (c) $9x^2 + \square + 4y^2$

Solution

(a) $6x = 2 \times x \times 3$ , so the last term is $3^2 = 9$ : $x^2 + 6x + 9 = (x + 3)^2\text{.}$

(b) $49 = 7^2$ , so the middle term is $\pm 2 \times x \times 7 = \pm 14x$ . The box is $14$ or $-14$ : $(x + 7)^2$ or $(x - 7)^2$ .

(a) $9$ ; (b) $14$ or $-14$ ; (c) $12xy$ or $-12xy$

Factorising to Calculate

You worked out differences of squares of numbers in lesson 3.6. Here the same work is seen as factorising: write $a^2 - b^2$ as the product $(a + b)(a - b)$ , then multiply.

Example 10: Numbers

Without a calculator, find (a) $57^2 - 43^2$ ; (b) $0.98^2 - 0.02^2$ .

Solution

(a) $57^2 - 43^2 = (57 + 43)(57 - 43) = 100 \times 14 = 1400$

(b) $0.98^2 - 0.02^2 = (0.98 + 0.02)(0.98 - 0.02) = 1 \times 0.96 = 0.96$

(a) $1400$ ; (b) $0.96$

Example 11: Without finding $a$ and $b$

Given that $a + b = 10$ and $a - b = 4$ , find $a^2 - b^2$ .

Solution

$a^2 - b^2 = (a + b)(a - b) = 10 \times 4 = 40$

Check: $a = 7$ and $b = 3$ fit, and $49 - 9 = 40$ . ✓

$a^2 - b^2 = 40$

Example 12: Challenge

Find the positive integers $x$ and $y$ for which $x^2 - 9y^2 = 13$ .

Solution

Factorise: $x^2 - (3y)^2 = (x + 3y)(x - 3y) = 13\text{.}$

$x + 3y$ is positive, because $x$ and $y$ are. The product $13$ is positive, so $x - 3y$ is positive too. And $x - 3y$ is less than $x + 3y$ .

$13$ is prime, so the only way to write it as a product of two positive integers, the smaller first, is $1 \times 13$ . So $x - 3y = 1$ and $x + 3y = 13$ .

Adding the two equations: $2x = 14$ , so $x = 7$ . Then $3y = 13 - 7 = 6$ , so $y = 2$ .

Check: $7^2 - 9(2^2) = 49 - 36 = 13\text{.}$ ✓

$x = 7$ , $y = 2$

Common Mistakes

Factorising a sum of squares: $x^2 + 9 \ne (x + 3)(x - 3)\text{,}$ which expands to $x^2 - 9$ .
Taking the square root of the number only: $25x^2 - 16 = (5x + 4)(5x - 4)\text{,}$ not $(25x + 4)(25x - 4)$ .
Calling three terms a perfect square without checking the middle term: $x^2 + 6x + 4$ is not $(x + 2)^2$ , which is $x^2 + 4x + 4$ .
Losing the sign: $4a^2 - 12ab + 9b^2 = (2a - 3b)^2\text{,}$ not $(2a + 3b)^2$ .
Testing the middle term with the whole terms instead of their square roots: for $x^2 + 10x + 25$ the test is $2 \times x \times 5$ , not $2 \times x^2 \times 25$ .

Practice

Try each question before you open its solution. Check each answer by expanding it.

Question 1

Factorise (a) $x^2 - 49$ ; (b) $16y^2 - 1$ ; (c) $36 - a^2$ .

ShowHide solution

(a) $x^2 - 7^2 = (x + 7)(x - 7)$

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

(a) $(x + 7)(x - 7)$ ; (b) $(4y + 1)(4y - 1)$ ; (c) $(6 + a)(6 - a)$

Question 2

Factorise (a) $4p^2 - 81q^2$ ; (b) $x^2y^2 - 25$ ; (c) $1 - 100k^2$ .

ShowHide solution

(a) $(2p)^2 - (9q)^2 = (2p + 9q)(2p - 9q)$

(b) $(xy)^2 - 5^2 = (xy + 5)(xy - 5)$

(a) $(2p + 9q)(2p - 9q)$ ; (b) $(xy + 5)(xy - 5)$ ; (c) $(1 + 10k)(1 - 10k)$

Question 3

Factorise (a) $(x + 2)^2 - 16$ ; (b) $9 - (y + 1)^2$ ; (c) $(x - 3)^2 - y^2$ .

ShowHide solution

(a) $\big((x + 2) + 4\big)\big((x + 2) - 4\big) = (x + 6)(x - 2)$

(b) $\big(3 + (y + 1)\big)\big(3 - (y + 1)\big) = (y + 4)(2 - y)\text{.}$ The bracket after the minus: $3 - y - 1 = 2 - y$ .

(a) $(x + 6)(x - 2)$ ; (b) $(y + 4)(2 - y)$ ; (c) $(x + y - 3)(x - y - 3)$

Question 4

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

ShowHide solution

(a) $\frac{1}{9} = (\frac{1}{3})^2$ : $(x + \frac{1}{3})(x - \frac{1}{3})$

(b) $\frac{1}{4} = (\frac{1}{2})^2$ and $2 \times x \times \frac{1}{2} = x$ . ✓ So $(x + \frac{1}{2})^2$ .

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

Question 5

Factorise (a) $x^2 + 14x + 49$ ; (b) $y^2 - 12y + 36$ ; (c) $25a^2 + 20ab + 4b^2$ .

ShowHide solution

(a) $49 = 7^2$ , $2 \times x \times 7 = 14x$ : $(x + 7)^2$

(b) $36 = 6^2$ , $2 \times y \times 6 = 12y$ , sign $-$ : $(y - 6)^2$

(a) $(x + 7)^2$ ; (b) $(y - 6)^2$ ; (c) $(5a + 2b)^2$

Question 6

Factorise (a) $9 - 6x + x^2$ ; (b) $1 + 16p^2 - 8p$ .

ShowHide solution

(a) $3^2$ and $x^2$ ; $2 \times 3 \times x = 6x$ , sign $-$ : $(3 - x)^2$ , which equals $(x - 3)^2$ .

(b) In descending powers: $16p^2 - 8p + 1$ . $(4p)^2$ and $1^2$ ; $2 \times 4p \times 1 = 8p$ , sign $-$ : $(4p - 1)^2$ .

(a) $(x - 3)^2$ (or $(3 - x)^2$ ); (b) $(4p - 1)^2$

Question 7

Factorise with an identity where you can. Where you cannot, say why. (a) $x^2 + 18x + 81$ ; (b) $x^2 + 16$ ; (c) $x^2 + 8x + 9$ ; (d) $9m^2 - 4n^2$

ShowHide solution

(a) $81 = 9^2$ and $2 \times x \times 9 = 18x$ : $(x + 9)^2$ .

(b) A sum of two squares: it cannot be factorised into two brackets.

(c) $9 = 3^2$ , but $2 \times x \times 3 = 6x$ , not $8x$ : not a perfect square, and three terms cannot be a difference of two squares.

(d) $(3m)^2 - (2n)^2 = (3m + 2n)(3m - 2n)\text{.}$

(a) $(x + 9)^2$ ; (b) no identity fits; (c) no identity fits; (d) $(3m + 2n)(3m - 2n)$

Question 8

Find the missing term or number that makes each a perfect square. (a) $x^2 - 10x + \square$ ; (b) $4x^2 + \square x + 9$ ; (c) $\square + 40y + 16$

ShowHide solution

(a) $10x = 2 \times x \times 5$ , so the last term is $25$ .

(b) $\pm 2 \times 2x \times 3 = \pm 12x\text{:}$ the box is $12$ or $-12$ .

(a) $25$ ; (b) $12$ or $-12$ ; (c) $25y^2$

Question 9

Without a calculator, find (a) $88^2 - 12^2$ ; (b) $7.5^2 - 2.5^2$ ; (c) $3.7^2 - 1.3^2$ .

ShowHide solution

(a) $(88 + 12)(88 - 12) = 100 \times 76 = 7600$

(b) $(7.5 + 2.5)(7.5 - 2.5) = 10 \times 5 = 50$

(a) $7600$ ; (b) $50$ ; (c) $12$

Question 10

(a) Given that $a + b = 12$ and $a - b = 3$ , find $a^2 - b^2$ . (b) Given that $x^2 - y^2 = 48$ and $x + y = 12$ , find $x - y$ , and then $x$ and $y$ .

ShowHide solution

(a) $a^2 - b^2 = (a + b)(a - b) = 12 \times 3 = 36\text{.}$

(b) $(x + y)(x - y) = 48$ , so $12(x - y) = 48$ and $x - y = 4$ .

Adding $x + y = 12$ and $x - y = 4$ : $2x = 16$ , $x = 8$ , and $y = 4$ . Check: $64 - 16 = 48$ . ✓

(a) $36$ ; (b) $x - y = 4$ , $x = 8$ , $y = 4$

Question 11

A square field has sides of $65.5$ m. In it is a square pond with sides of $34.5$ m. Without a calculator, find the area of the field not covered by the pond.

ShowHide solution

Area $= 65.5^2 - 34.5^2 = (65.5 + 34.5)(65.5 - 34.5) = 100 \times 31 = 3100$

$3100$ m²

Question 12

Challenge. Given that $x$ and $y$ are positive integers, find all the solutions of $x^2 - 4y^2 = 21$ .

ShowHide solution

$(x + 2y)(x - 2y) = 21\text{.}$ Since $x + 2y$ and the product are positive, $x - 2y$ is positive, and it is the smaller factor.

$21 = 1 \times 21$ or $3 \times 7$ .

$x - 2y = 1$ and $x + 2y = 21$ : $2x = 22$ , $x = 11$ , $4y = 20$ , $y = 5$ .

$x - 2y = 3$ and $x + 2y = 7$ : $2x = 10$ , $x = 5$ , $4y = 4$ , $y = 1$ .

Check: $121 - 4(25) = 21$ and $25 - 4(1) = 21$ . ✓

$x = 11$ , $y = 5$ ; or $x = 5$ , $y = 1$

Key Points

$a^2 - b^2 = (a + b)(a - b)\text{:}$ two squares with a minus between them.
$a^2 + 2ab + b^2 = (a + b)^2$ and $a^2 - 2ab + b^2 = (a - b)^2\text{:}$ first and last terms squares with a $+$ sign, middle term twice the product of their square roots.
Ask "what was squared?" to find $a$ and $b$ ; a bracket can be one of the squares.
A sum of two squares, such as $x^2 + 9$ , cannot be factorised into two brackets.
$(4 - x)^2 = (x - 4)^2$ .
Check every factorisation by expanding it back.

Factorising Using the Identities

Before you start

The Identities Read Backwards

Factorising with the identities

Recognising a Difference of Two Squares

A difference of two squares has

Example 1: Squares of a letter and a number

Example 2: The number first, and two letters

Example 3: A bracket as one of the squares

Example 4: A fraction

Recognising a Perfect Square

A perfect square has

Example 5: Testing and factorising

Example 6: The number first

Example 7: A fraction

When an Identity Does Not Fit

Example 8: Check before you factorise

A common factor can still come out

The Missing Term of a Perfect Square

Example 9: The missing term

Factorising to Calculate

Example 10: Numbers

Example 11: Without finding aaa and bbb

Example 12: Challenge

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 11: Without finding $a$ and $b$