Factoring Special Cases

Some expressions have a special shape that lets you factor them in one step, if you recognize it. The two you need are the difference of squares, like $x^2 - 49$, and the perfect-square trinomial, like $x^2 + 10x + 25$. This page also pulls together everything in the unit into one strategy, so that you can look at any Grade 10 expression and know where to start.

Key ideas

Difference of squares

From polynomial operations, you know that $(a + b)(a - b) = a^2 - b^2$: the middle terms cancel. Read backwards, that’s a factoring rule:

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

To use it, the expression must have:

• exactly two terms,
• both of them perfect squares,
• a minus sign between them.

Find what each term is the square of, then write one bracket with $+$ and one with $-$. For example, $9x^2 - 16 = (3x)^2 - 4^2 = (3x + 4)(3x - 4)$.

Cutting the leftover shape into two pieces and rearranging them shows that $a^2 - b^2 = (a + b)(a - b)$.

A sum of squares doesn’t factor

$x^2 + 9$ is a sum of squares, and it does not factor over the real numbers. Check the tempting guesses: $(x + 3)(x - 3) = x^2 - 9$ and $(x + 3)^2 = x^2 + 6x + 9$. Neither one is $x^2 + 9$.

Perfect-square trinomials

Squaring a binomial gives a trinomial with a special pattern:

$$\begin{aligned} a^2 + 2ab + b^2 &= (a + b)^2 \\ a^2 - 2ab + b^2 &= (a - b)^2 \end{aligned}$$

A trinomial is a perfect square if:

1. the first and last terms are perfect squares, $a^2$ and $b^2$ (and the last term is positive), and
2. the middle term is twice the product of $a$ and $b$, that is $\pm 2ab$.

The sign of the middle term tells you the sign in the bracket. For $4x^2 - 12x + 9$: $4x^2 = (2x)^2$, $9 = 3^2$, and $2(2x)(3) = 12x$ matches the middle term, so $4x^2 - 12x + 9 = (2x - 3)^2$.

You can always factor a perfect-square trinomial with the methods for trinomials instead; recognizing the pattern just saves time.

Factoring in two steps

Sometimes one method leaves something that can be factored again. Always take out a common factor first, then check every factor to see whether it can be broken down further:

$$2x^2 - 18 = 2(x^2 - 9) = 2(x + 3)(x - 3)$$ $$x^4 - 16 = (x^2 + 4)(x^2 - 4) = (x^2 + 4)(x + 2)(x - 2)$$

An expression is fully factored when none of its factors can be factored any further.

A factoring strategy

Use these steps for every expression in this unit:

Step	Ask yourself	Method
1	Do all the terms share a factor?	Take out the common factor. If the first term is negative, take out a negative.
2	Two terms?	Is it a difference of squares, $a^2 - b^2$? A sum of squares doesn’t factor.
3	Three terms, starting with $x^2$?	Find two numbers with product $c$ and sum $b$. Look out for a perfect square.
4	Three terms, starting with $ax^2$, $a \ne 1$?	Decomposition (product $ac$, sum $b$) or inspection. Look out for a perfect square.
5	Four terms?	Factor by grouping.
6	Can any factor be factored again?	Repeat. Then check by expanding.

Some expressions don’t factor

Not every expression factors over the integers. For example:

• $x^2 + 4$ is a sum of squares;
• $x^2 + 3x + 1$ has no integer pair with product $1$ and sum $3$;
• $x^2 - 7$ is a difference, but $7$ isn’t a perfect square.

If you’ve tried every step and nothing works, say that the expression doesn’t factor. That’s a complete answer. (Later, the quadratic formula will let you solve equations with expressions like these.)

Worked examples

Example 1: Differences of squares

Factor.

• (a) $x^2 - 49$
• (b) $9x^2 - 16$
• (c) $36a^2 - b^2$

Solution. Write each term as a square, then use $a^2 - b^2 = (a + b)(a - b)$.

(a) $x^2 - 49 = x^2 - 7^2 = (x + 7)(x - 7)$

(b) $9x^2 - 16 = (3x)^2 - 4^2 = (3x + 4)(3x - 4)$

(c) $36a^2 - b^2 = (6a)^2 - b^2 = (6a + b)(6a - b)$

Check (b): $(3x + 4)(3x - 4) = 9x^2 - 12x + 12x - 16 = 9x^2 - 16$. ✓

Example 2: Is it a perfect square?

Decide whether each trinomial is a perfect square, then factor it.

• (a) $x^2 + 10x + 25$
• (b) $4x^2 - 12x + 9$
• (c) $9x^2 + 15x + 4$

Solution.

(a) $x^2 = (x)^2$ and $25 = 5^2$. Twice the product is $2(x)(5) = 10x$, which matches. So $x^2 + 10x + 25 = (x + 5)^2$.

(b) $4x^2 = (2x)^2$ and $9 = 3^2$. Twice the product is $2(2x)(3) = 12x$, and the middle term is $-12x$. So $4x^2 - 12x + 9 = (2x - 3)^2$.

(c) $9x^2 = (3x)^2$ and $4 = 2^2$, but twice the product is $2(3x)(2) = 12x$, not $15x$. So it’s not a perfect square. Use decomposition instead: $ac = 36$, sum $15$, so the integers are $3$ and $12$.

$$\begin{aligned} 9x^2 + 15x + 4 &= 9x^2 + 3x + 12x + 4 \\ &= 3x(3x + 1) + 4(3x + 1) \\ &= (3x + 1)(3x + 4) \end{aligned}$$

Check (c): $(3x + 1)(3x + 4) = 9x^2 + 12x + 3x + 4 = 9x^2 + 15x + 4$. ✓

Example 3: Two steps

Factor fully.

• (a) $2x^2 - 18$
• (b) $5x^2 + 20x + 20$
• (c) $x^4 - 81$

Solution.

(a) Neither $2x^2$ nor $18$ is a perfect square, but take out the common factor $2$ first and a difference of squares appears:

$$2x^2 - 18 = 2(x^2 - 9) = 2(x + 3)(x - 3)$$

(b) Take out $5$, then spot a perfect square ($2(x)(2) = 4x$ ✓):

$$5x^2 + 20x + 20 = 5(x^2 + 4x + 4) = 5(x + 2)^2$$

(c) $x^4 = (x^2)^2$ and $81 = 9^2$, so this is a difference of squares. Then one of the factors is a difference of squares again:

$$\begin{aligned} x^4 - 81 &= (x^2 + 9)(x^2 - 9) \\ &= (x^2 + 9)(x + 3)(x - 3) \end{aligned}$$

$x^2 + 9$ is a sum of squares, so it doesn’t factor. We’re done.

Example 4: Using the strategy

Factor fully, or say that the expression doesn’t factor.

• (a) $3x^2 + 12$
• (b) $6x^2 + 7x - 3$
• (c) $x^3 - 2x^2 - 9x + 18$
• (d) $x^2 + x + 3$

Solution.

(a) Step 1: common factor $3$, giving $3(x^2 + 4)$. Step 2: $x^2 + 4$ is a sum of squares, which doesn’t factor. The answer is $3(x^2 + 4)$.

(b) No common factor. Three terms with $a = 6$: use decomposition. $ac = -18$, sum $7$, so the integers are $9$ and $-2$.

$$\begin{aligned} 6x^2 + 7x - 3 &= 6x^2 + 9x - 2x - 3 \\ &= 3x(2x + 3) - 1(2x + 3) \\ &= (2x + 3)(3x - 1) \end{aligned}$$

(c) Four terms: group them. Then the bracket $x^2 - 9$ factors again.

$$\begin{aligned} x^3 - 2x^2 - 9x + 18 &= x^2(x - 2) - 9(x - 2) \\ &= (x - 2)(x^2 - 9) \\ &= (x - 2)(x + 3)(x - 3) \end{aligned}$$

(d) No common factor. We need product $3$ and sum $1$. The pairs $1, 3$ and $-1, -3$ have sums $4$ and $-4$, so $x^2 + x + 3$ doesn’t factor over the integers.

Common mistakes

Factoring a sum of squares. $x^2 + 25$ is not $(x + 5)(x - 5)$ (that’s $x^2 - 25$) and it’s not $(x + 5)^2$ (that’s $x^2 + 10x + 25$). A sum of squares doesn’t factor.

Assuming a perfect square without checking the middle term. $9x^2 + 15x + 4$ starts and ends with perfect squares, but $(3x + 2)^2 = 9x^2 + 12x + 4$. Always check that the middle term is $2ab$.

Stopping too early. $x^4 - 16 = (x^2 + 4)(x^2 - 4)$ isn’t finished, because $x^2 - 4$ is another difference of squares. Likewise, $(2x + 6)(x - 3)$ isn’t fully factored, because $2x + 6 = 2(x + 3)$. Check every factor.

Forcing a difference of squares. $x^2 - 8$ isn’t a difference of squares over the integers, because $8$ isn’t a perfect square. Don’t write $(x + 4)(x - 4)$ or $(x + 2)(x - 4)$; expand to see that they don’t work.

Missing a common factor that hides the pattern. $3x^2 - 75$ doesn’t look like a difference of squares until you take out the $3$: $3(x^2 - 25) = 3(x + 5)(x - 5)$.

Getting the order wrong when the constant comes first. $25 - y^2 = (5 + y)(5 - y)$. It’s not $(y + 5)(y - 5)$, which equals $y^2 - 25$, the opposite.

Practice

1. (Warm-up) Factor.

• (a) $x^2 - 64$
• (b) $25 - y^2$
• (c) $4x^2 - 81$

Solution

(a) $(x + 8)(x - 8)$

(b) $25 - y^2 = 5^2 - y^2 = (5 + y)(5 - y)$

(c) $4x^2 - 81 = (2x)^2 - 9^2 = (2x + 9)(2x - 9)$

2. (Warm-up) Decide whether each is a perfect-square trinomial. Then factor it.

• (a) $x^2 + 14x + 49$
• (b) $x^2 - 12x + 36$
• (c) $x^2 + 8x + 12$

Solution

(a) Yes: $2(x)(7) = 14x$. So $(x + 7)^2$.

(b) Yes: $2(x)(6) = 12x$ and the middle term is negative. So $(x - 6)^2$.

(c) No: $12$ isn’t a perfect square. Product $12$, sum $8$: $2$ and $6$. So $(x + 2)(x + 6)$.

3. (Core) Factor $16x^2 - 40x + 25$.

Solution

$16x^2 = (4x)^2$ and $25 = 5^2$. Twice the product is $2(4x)(5) = 40x$, which matches the middle term (with a minus sign). So

$$16x^2 - 40x + 25 = (4x - 5)^2$$

Check: $(4x - 5)^2 = 16x^2 - 40x + 25$. ✓

4. (Core) Factor fully.

• (a) $3x^2 - 75$
• (b) $50 - 2a^2$

Solution

(a)

$$3x^2 - 75 = 3(x^2 - 25) = 3(x + 5)(x - 5)$$

(b)

$$50 - 2a^2 = 2(25 - a^2) = 2(5 + a)(5 - a)$$

5. (Core) Factor $49x^2 - 100y^2$.

Solution

$$49x^2 - 100y^2 = (7x)^2 - (10y)^2 = (7x + 10y)(7x - 10y)$$

6. (Core) Factor fully, or say that the expression doesn’t factor.

• (a) $4x^2 + 36$
• (b) $2x^2 + 16x + 32$
• (c) $x^4 - 1$
• (d) $x^2 - 5x + 9$

Solution

(a) Take out $4$: $4(x^2 + 9)$. The sum of squares $x^2 + 9$ doesn’t factor, so the answer is $4(x^2 + 9)$.

(b) Take out $2$, then spot a perfect square:

$$2x^2 + 16x + 32 = 2(x^2 + 8x + 16) = 2(x + 4)^2$$

(c) Difference of squares twice:

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

(d) We need product $9$ and sum $-5$. The pairs $1, 9$, $3, 3$, $-1, -9$ and $-3, -3$ have sums $10$, $6$, $-10$ and $-6$. None is $-5$, so $x^2 - 5x + 9$ doesn’t factor over the integers.

7. (Core) A square lawn has sides of $x$ metres. A square flower bed with sides of $3$ m is dug out of one corner.

• (a) Write the remaining lawn area as an expression, and factor it.
• (b) Find the remaining area when $x = 10$, using both forms.

Solution

(a) The remaining area is $x^2 - 9$ square metres, which factors as $(x + 3)(x - 3)$.

(b) Using $x^2 - 9$: $10^2 - 9 = 91$ m². Using the factored form: $(13)(7) = 91$ m². ✓

8. (Challenge) Use the difference of squares to calculate these without a calculator.

• (a) $47 \times 53$
• (b) $103^2 - 97^2$

Solution

(a) $47 = 50 - 3$ and $53 = 50 + 3$, so

$$47 \times 53 = (50 + 3)(50 - 3) = 50^2 - 3^2 = 2500 - 9 = 2491$$

(b)

$$103^2 - 97^2 = (103 + 97)(103 - 97) = 200 \times 6 = 1200$$

9. (Challenge)

• (a) Find all values of $k$ so that $4x^2 + kx + 25$ is a perfect-square trinomial.
• (b) Factor fully: $x^4 - 13x^2 + 36$.

Solution

(a) $4x^2 = (2x)^2$ and $25 = 5^2$, so the middle term must be $\pm 2(2x)(5) = \pm 20x$. So $k = 20$ or $k = -20$, giving $(2x + 5)^2$ or $(2x - 5)^2$.

(b) Treat $x^2$ like a single variable: we need product $36$ and sum $-13$, which are $-4$ and $-9$.

$$\begin{aligned} x^4 - 13x^2 + 36 &= (x^2 - 4)(x^2 - 9) \\ &= (x + 2)(x - 2)(x + 3)(x - 3) \end{aligned}$$

Check with $x = 1$: the original is $1 - 13 + 36 = 24$, and $(3)(-1)(4)(-2) = 24$. ✓