MPM2D Perfect Squares and Difference of Squares: Notes and Practice

Here is a trick for multiplying $49 \times 51$ in your head. Write the numbers as $50 - 1$ and $50 + 1$ . A sum times a difference of the same two numbers loses its middle terms: $(50 - 1) (50 + 1) = 5 0^{2} - 1^{2} = 2500 - 1 = 2499 .$ In the same way, $2 1^{2} = (20 + 1)^{2} = 400 + 40 + 1 = 441 .$ Both tricks use the special products from the first lesson. This lesson turns them around: when you can recognize an expression as one of these special patterns, you can factor it in one step, without searching through factor pairs.

Difference of squares

a^{2} - b^{2} = (a - b) (a + b)

Expanding the right side shows why: $(a - b) (a + b) = a^{2} + ab - ab - b^{2} = a^{2} - b^{2} .$ The middle terms are opposites, so they cancel. To use the pattern, the expression must have exactly two terms, both must be perfect squares, and they must be subtracted. A perfect square is something that equals an expression times itself: $25 = 5^{2}$ , $4 x^{2} = (2 x)^{2}$ , $81 y^{2} = (9 y)^{2}$ . Find $a$ and $b$ by taking the square root of each term, then write one bracket with $-$ and one with $+$ . For example, $x^{2} - 49 = (x - 7) (x + 7)$ and $4 x^{2} - 25 = (2 x)^{2} - 5^{2} = (2 x - 5) (2 x + 5) .$

Fig 1A square of side a with a b by b corner removed has area a² − b². Cutting off the bottom strip and moving it to the side makes one rectangle a + b long and a − b wide.

A sum of squares is different

x^{2} + 25

is a sum of two squares, not a difference. Brackets like

(x + 5) (x + 5)

give

x^{2} + 10 x + 25

and

(x + 5) (x - 5)

gives

x^{2} - 25

, so no pair of binomials with integer numbers works. In fact a second-degree sum of squares such as

x^{2} + 25

4 x^{2} + 9

has no binomial factors using any real numbers at all. (A common factor can still come out:

2 x^{2} + 8 = 2 (x^{2} + 4)

, and the bracket stops there.)

VideoSee why a sum times a difference loses its middle terms, and how to spot a difference of squares to factor it.Watch on YouTube

Perfect square trinomials

a^{2} + 2 ab + b^{2} = (a + b)^{2} a^{2} - 2 ab + b^{2} = (a - b)^{2}

A perfect square trinomial is the result of squaring a binomial. You can recognize one in three checks: the first term is a perfect square ( $a^{2}$ ), the last term is a perfect square ( $b^{2}$ ) and positive, and the middle term is twice the product of the square roots ( $2 ab$ ), with either sign. The sign of the middle term becomes the sign in the bracket. Test $4 x^{2} - 12 x + 9$ : $4 x^{2} = (2 x)^{2}$ , $9 = 3^{2}$ , and $2 (2 x) (3) = 12 x$ , which matches the middle term. So $4 x^{2} - 12 x + 9 = (2 x - 3)^{2}$ . Now test $x^{2} + 10 x + 16$ : $16 = 4^{2}$ , but $2 (x) (4) = 8 x$ , not $10 x$ , so it is not a perfect square; it factors as $(x + 2) (x + 8)$ instead. The product and sum method from the last lesson also works on a perfect square: for $x^{2} + 10 x + 25$ the pair is $5$ and $5$ , giving $(x + 5) (x + 5) = (x + 5)^{2}$ . Recognizing the pattern just gets you there faster.

Expression	Squares	Middle term test	Factored
$x^{2} + 10 x + 25$	$x^{2}$ and $5^{2}$	$2 (x) (5) = 10 x$ ✓	$(x + 5)^{2}$
$4 x^{2} - 12 x + 9$	$(2 x)^{2}$ and $3^{2}$	$2 (2 x) (3) = 12 x$ ✓	$(2 x - 3)^{2}$
$9 x^{2} + 30 x y + 25 y^{2}$	$(3 x)^{2}$ and $(5 y)^{2}$	$2 (3 x) (5 y) = 30 x y$ ✓	$(3 x + 5 y)^{2}$
$x^{2} + 10 x + 16$	$x^{2}$ and $4^{2}$	$2 (x) (4) = 8 x$ , not $10 x$	not a perfect square: $(x + 2) (x + 8)$

Testing for a perfect square trinomial

VideoLearn to recognize a perfect square trinomial and write it as the square of a binomial.Watch on YouTube

Factoring fully: one strategy for every expression

Common factor first. Take out the GCF (and $- 1$ if the leading term is negative). This often reveals a pattern: $2 x^{2} - 18 = 2 (x^{2} - 9)$ .
Count the terms in what is left. Two terms: is it a difference of squares? Three terms: is it a perfect square trinomial? If not, use product and sum, or decomposition when $a \neq = 1$ . Four terms: try grouping.
Look inside every bracket to see whether it can be factored again: $2 (x^{2} - 9) = 2 (x - 3) (x + 3)$ .
Check by expanding your answer back to the original.

The patterns work with any expression in the place of $a$ or $b$ , not just single letters. In $(x + 3)^{2} - 16$ , the first square is $(x + 3)^{2}$ and the second is $4^{2}$ , so $a = x + 3$ and $b = 4$ : $(x + 3)^{2} - 16 = (x + 3 - 4) (x + 3 + 4) = (x - 1) (x + 7) .$ Check by expanding both: $(x + 3)^{2} - 16 = x^{2} + 6 x + 9 - 16 = x^{2} + 6 x - 7,$ and $(x - 1) (x + 7) = x^{2} + 6 x - 7$ . Going beyond second-degree expressions, step 3 matters even more: $x^{4} - 16 = (x^{2} - 4) (x^{2} + 4) = (x - 2) (x + 2) (x^{2} + 4),$ because the first bracket is itself a difference of squares, while $x^{2} + 4$ is a sum of squares and stops there.

Squares worth knowing on sight

1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144

are the squares of 1 to 12. A term like

36 x^{2}

(6 x)^{2}

, and

x^{2} y^{2}

(x y)^{2}

. A coefficient that is not a perfect square, such as

2 x^{2}

8

, will not fit a square pattern until you take out a common factor.

Perfect Squares and Difference of Squares

Key things to remember

Difference of squares

Perfect square trinomials

Factoring fully: one strategy for every expression

Worked examples

Where marks get lost

Flashcards 8

Practice