 Hong Kong
Stage 4 - Stage 5

# Perfect squares

Lesson

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

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

$\left(a-b\right)^2=a^2-2ab+b^2$(ab)2=a22ab+b2

Notice a pattern in the expansion 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 expanded expression that fits this pattern, we can go backwards and factorise it.

That is:

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

There is a very similar result with the differences of two squares, where we used the result of expansion to go backwards and factorise.

#### Examples

##### Question 1

Which of the following are perfect square expansions?

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

Solution:

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

B) $x^2-4x+16$x24x+16

Solution:

We can write $x^2-4x+16$x24x+16 as $\left(x\right)^2-4x+\left(-4\right)^2$(x)24x+(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 expression.

C) $x^2-18x+81$x218x+81

Solution:

We can write $x^2-18x+81$x218x+81 as $\left(x\right)^2+2\left(-9x\right)+\left(-9\right)^2$(x)2+2(9x)+(9)2

We have the square of $x$x, the square of $\left(-9\right)$(9), and double the product of $x$x and $\left(-9\right)$(9). So this is a perfect square expression that can be factorised.

D) $9+12x+4x^2$9+12x+4x2

Solution:

We can write $9+12x+4x^2$9+12x+4x2 as $\left(3\right)^2+2\left(3\times2x\right)+\left(2x\right)^2$(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 expression that can be factorised.

##### Question 2

By expanding $\left(b+7\right)^2$(b+7)2, we want to determine the factorisation for $b^2+14b+49$b2+14b+49.

1. First expand $\left(b+7\right)^2$(b+7)2.

2. Hence, factorise $b^2+14b+49$b2+14b+49.

##### Question 3

Factorise the expression $36y^2-12yz+z^2$36y212yz+z2

Think about how to attain $a$a and $b$b, and how to check if this really is a perfect square expression.

Do

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

##### Question 4

Factorise $x^2-12x+36$x212x+36.

##### Question 5

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

Think about how to make this a perfect square expression by first factorising all the coefficients.

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 an HCF of $3$3, so let's factorise 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 brackets is a perfect square expression where $\sqrt{p^2}=p$p2=p and $\sqrt{4}=2$4=2.

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

Watch Out

Always see if you can factorise expressions using the most basic method of $AB+AC=A\left(B+C\right)$AB+AC=A(B+C) first if you can. It'll make things a lot easier later on!

##### Question 6

Factorise $-18b^2-12ab-2a^2$18b212ab2a2.