2.07 Factoring with special products

We have looked at factoring trinomials of the form $ax^2+bx+c$ax2+bx+c using grouping or other strategies. We can still use these strategies, but it is worth checking for patterns first to see if we can be more efficient.

 

Factoring a difference of two squares

Recall that if we distribute a sum and difference, we got a difference of two squares:

$\left(a+b\right)\left(a-b\right)=a^2-b^2$(a+b)(a−b)=a2−b2

We already know how to use the above property to distribute the parenthesized expression on the left, but now we can also use it to factor the one on the right, since factoring is the opposite of distribution.

 

Worked example

Question 1

Factor the expression $16x^2-81y^2$16x2−81y2.

Think: We are looking for expressions of the form $a^2-b^2$a2−b2, what would that make $a$a and $b$b in this scenario? $a$a and $b$b are the square roots of $a^2$a2 and $b^2$b2 respectively.

Do: 

$\sqrt{16x^2}$√16x2	$=$=	$\sqrt{16}\sqrt{x^2}$√16√x2
	$=$=	$4x$4x
$\sqrt{81y^2}$√81y2	$=$=	$\sqrt{81}\sqrt{y^2}$√81√y2
	$=$=	$9y$9y
Therefore
$\left(4x\right)^2$(4x)2	$=$=	$16x^2$16x2
and
$\left(9y\right)^2$(9y)2	$=$=	$81y^2$81y2
So
$16x^2-81y^2$16x2−81y2	$=$=	$\left(4x\right)^2-\left(9y\right)^2$(4x)2−(9y)2
		$\left(4x+9y\right)\left(4x-9y\right)$(4x+9y)(4x−9y)

 

Practice questions

Question 2

Question 3

Question 4

 

Perfect square trinomials

Another type of binomial expression that can be factored very nicely are squares of binomials. We've already come across some of these and know how to distribute them:

$\left(a+b\right)^2=a^2+2ab+b^2$(a+b)2=a2+2ab+b2

$\left(a-b\right)^2=a^2-2ab+b^2$(a−b)2=a2−2ab+b2

Notice a pattern in the distribution of $\left(a+b\right)^2$(a+b)2:

$a^2$a2	$+$+	$2ab$2ab	$+$+	$b^2$b2
$\uparrow$↑		$\uparrow$↑		$\uparrow$↑
Square of the first term		Double the product of the two terms		Square of the second term

 

So if we have an distributed expression that fits this pattern, we can go backwards and factor it.

That is:

$a^2+2ab+b^2=\left(a+b\right)^2$a2+2ab+b2=(a+b)2

Worked examples

Question 5

Which of the following are perfect square trinomials?

A) $x^2+6x+9$x2+6x+9

Solution:

We can write $x^2+6x+9$x2+6x+9 as $\left(x\right)^2+2\times\left(3\right)\left(x\right)+\left(3\right)^2$(x)2+2×(3)(x)+(3)2

We have the square of $x$x, the square of $3$3, and double the product of $x$x and $3$3. So this is a perfect square expression that can be factored using this pattern.

 

B) $x^2-4x+16$x2−4x+16

Solution:

We can write $x^2-4x+16$x2−4x+16 as $\left(x\right)^2-\left(4\right)\left(x\right)+\left(-4\right)^2$(x)2−(4)(x)+(−4)2

We have the square of $x$x and the square of $\left(-4\right)$(−4). But the other term, $-4x$−4x is not double the product of $x$x and $\left(-4\right)$(−4). So this is not a perfect square trinomial.

 

C) $9-12x+4x^2$9−12x+4x2

Solution:

We can write $9-12x+4x^2$9−12x+4x2 as $\left(3\right)^2-2\times\left(3\right)\left(2x\right)+\left(2x\right)^2$(3)2−2×(3)(2x)+(2x)2

We have the square of $3$3, the square of $2x$2x, and double the product of $3$3 and $2x$2x. So this is a perfect square trinomial that can be factored.

 

Question 6

Factor the expression $36y^2-12y+1^2$36y2−12y+12

Think: How do we obtain $a$a and $b$b, and check if this really is a perfect square trinomial?

Do:

$\sqrt{36y^2}$√36y2	$=$=	$\sqrt{36}\sqrt{y^2}$√36√y2
	$=$=	$6y$6y
$\sqrt{1^2}$√12	$=$=	$1$1
Therefore $a$a is $6y$6y and our $b$b is $1$1.
$2\times a\times b$2×a×b	$=$=	$2\times6y\times1$2×6y×1
	$=$=	$12y$12y
which is the same as the middle term so this is a perfect square trinomial
So
$36y^2-12yz+z^2$36y2−12yz+z2	$=$=	$\left(6y\right)^2-2\left(6y\right)\times\left(1\right)+1^2$(6y)2−2(6y)×(1)+12
		$\left(6y-1\right)^2$(6y−1)2

 

Question 7

Factor the following expression completely: $3p^2+12p+12$3p2+12p+12

Think: Any time we are asked to factor, we should first check for a GCF. Notice, there is a common factor of $3$3 across all terms.

Do: We can see here that $3p^2$3p2 and $12$12 are not perfect squares! So how can we find $a$a and $b$b?

If we look at this another way we can see that the coefficients of all three terms have a GCF of $3$3, so let's factor that out first.

$3p^2+12p+12=3\left(p^2+4p+4\right)$3p2+12p+12=3(p2+4p+4)

Now the expression in the parentheses is a perfect square trinomial where $\sqrt{p^2}=p$√p2=p and $\sqrt{4}=2$√4=2.

Therefore

$3p^2+12p+12$3p2+12p+12	$=$=	$3\left(p^2+4p+4\right)$3(p2+4p+4)
	$=$=	$3\left(p^2+2\times2p+2^2\right)$3(p2+2×2p+22)
	$=$=	$3\left(p+2\right)^2$3(p+2)2

 

 

Practice questions

Question 8

Question 9

Question 10