Sometimes we come across quadratics that are impossible to factorise with what we know so far, but we can use our knowledge of binomial squares to factorise part of them and simplify the expression. This technique is called completing the square. Let's start by having a look at how it works in the following video:
So to review, we already know that binomial squares follow the expansion formula $\left(a+b\right)^2=a^2+2ab+b^2$(a+b)2=a2+2ab+b2. We're going to be looking at factorisation examples of the form:
$x^2+2ax+a^2=\left(x+a\right)^2$x2+2ax+a2=(x+a)2
The key is to note that in the original quadratic, the third term is always the square of half the coefficient of $x$x. The coefficient halved also helps give us the factorised form. For example in the perfect square $x^2+6x+9$x2+6x+9, halving the $x$x coefficient $6$6 gives us $3$3, which gives us the third term $9$9 when squared. That means we can factorise the quadratic to $\left(x+3\right)^2$(x+3)2. This is another way to look at the relationship shown with a drawn square in the video. We can use this knowledge to factorise irregular monic quadratics into the form $\left(x+a\right)^2+b$(x+a)2+b, where a and b are any numbers.
For example, $x^2+8x+10$x2+8x+10 is not a perfect square, but we can use part of it to form one. Looking at just the first two terms we have $x^2+8x$x2+8x. If this was part of a perfect square what would be the third term?
Let's follow our previous technique by halving and then squaring the $x$x coefficient. $\left(8\div2\right)^2=4^2$(8÷2)2=42 $=$= $16$16, which should be the third term.
Following in the footsteps of the video example, we can then rewrite
$x^2+8x+10$x2+8x+10 | $=$= | $\left(x^2+8x+16\right)+10-16$(x2+8x+16)+10−16 |
$=$= | $\left(x+4\right)^2+10-16$(x+4)2+10−16 | |
$=$= | $\left(x+4\right)^2-6$(x+4)2−6 |
Wow, we managed to factorise one part of a quadratic! Doesn't it look much simpler?
Let's work through some more examples:
What coefficient of $x$x would make the following a perfect square?
$x^2+\editable{}x+36$x2+x+36
Think about the relationship between the second and third terms in the equation $\left(x+a\right)^2=x^2+2ax+a^2$(x+a)2=x2+2ax+a2
Do We know that we can halve the $x$x coefficient and then square it to get to the third term.
That means we can square root the third term and double it to get to the $x$x coefficient.
$2\times\sqrt{36}$2×√36 | $=$= | $2\times6$2×6 |
$=$= | $12$12 |
The coefficient must then be $12$12.
Complete the square by finding $j$j and $k$k:
$x^2-14x+j=\left(x+k\right)^2$x2−14x+j=(x+k)2
Think about how $k$k is related to the $x$x coefficient
Do
$j$j | $=$= | $\left(\left(-14\right)\div2\right)^2$((−14)÷2)2 |
$=$= | $\left(-7\right)^2$(−7)2 | |
$=$= | $49$49 |
$k$k is half the $x$x coefficient which is $-14$−14, therefore $k=-7$k=−7.
Factorise the quadratic using the method of completing the square to get it into the form $\left(x+a\right)^2+b$(x+a)2+b
$x^2-10x+98$x2−10x+98
Think about leaving the third term aside and considering the first two first
Do: If we wanted $x^2-10x+k$x2−10x+k to be a perfect square then:
$k$k | $=$= | $\left(\left(-10\right)\div2\right)^2$((−10)÷2)2 |
$=$= | $\left(-5\right)^2$(−5)2 | |
$=$= | $25$25 |
That means
$x^2-10x+98$x2−10x+98 | $=$= | $\left(x^2-10x+25\right)+98-25$(x2−10x+25)+98−25 |
$=$= | $\left(x-5\right)^2+98-25$(x−5)2+98−25 | |
$=$= | $\left(x-5\right)^2+73$(x−5)2+73 |
Using the method of completing the square, rewrite $x^2+3x+6$x2+3x+6 in the form $\left(x+b\right)^2+c$(x+b)2+c.