Perfect Square Trinomials

It’s useful to think of “completing the square” as short for “completing the perfect square trinomial”.

First, recall that a perfect square is a number whose square root is an integer. The easiest way to think of perfect squares is to square the counting numbers:

$1^2, 2^2, 3^2, 4^2, 5^2, ...$

A perfect square trinomial is a polynomial whose square root is a binomial. The easiest way to think of perfect square trinomials is to square the binomials obtained by adding counting numbers to a variable $x$:

$(x+1)^2,$ $(x+2)^2,$ $(x+3)^2,$ $(x+4)^2,$ $(x+5)^2,$ $\vdots$

Recall that to square something is to multiply it by itself:

$[x + 1](x + 1),$ $[x + 2](x + 2),$ $[x + 3](x + 3),$ $[x + 4](x + 4),$ $[x + 5](x + 5),$ $\vdots$

Let’s distribute the second factor into the first factor:

$[ x(x+1) + 1(x+1) ],$ $[ x(x+2) + 2(x+2) ],$ $[ x(x+3) + 3(x+3) ],$ $[ x(x+4) + 4(x+4) ],$ $[ x(x+5) + 5(x+5) ],$ $\vdots$

Now further distribute:

$[ x^2 + 1x + 1x + 1 ],$ $[ x^2 + 2x + 2x + 4 ],$ $[ x^2 + 3x + 3x + 9 ],$ $[ x^2 + 4x + 4x + 16 ],$ $[ x^2 + 5x + 5x + 25 ],$ $\vdots$

Finally, combine like terms:

$x^2 + 2x + 1,$ $x^2 + 4x + 4,$ $x^2 + 6x + 9,$ $x^2 + 8x + 16,$ $x^2 + 10x + 25,$ $\vdots$

We see the general pattern for the square of a monomial of the form $x + b$ would then be

$x^2 + 2bx + b^2$

So, a perfect square trinomial must, by definition, be able to be written as the square of some binomial. We just saw that if we square any monomial of the form $x + b$, a perfect square trinomial of the form $x^2 + 2bx + b^2$ emerges. This tells us that if we run into a quadratic polynomial of the form $x^2 + 2bx$, we should add $b^2$ to complete the square. In words, we add the square of half the $x$ coefficient.

The monomial $x + b$ is pretty general, but $ax + b$ is even more general. Think of them as the same, but in the first one we assumed $a = 1$.

$\begin{alignedat}{1} (ax + b)^2 &= [ax + b](ax + b)\\ &= [ax(ax + b) + b(ax + b)]\\ &= [a^2x^2 + abx + abx + b^2]\\ &= a^2x^2 + 2abx + b^2\\ \end{alignedat}$

At this point, just remind yourself that if $a = 0$, the equation would reduce to $b^2 = b^2$. So let’s assume $a \neq 0$, which means $a^2 \neq 0$. We are now able to divide both sides by $a^2$:

$\begin{alignedat}{1} (ax + b)^2 &= a^2x^2 + 2abx + b^2\\ \frac{(ax + b)^2}{a^2} &= \frac{a^2x^2 + 2abx + b^2}{a^2}\\ \bigg(\frac{ax + b}{a}\bigg)^2 &= \frac{a^2x^2}{a^2} + \frac{2abx}{a^2} + \frac{b^2}{a^2}\\ \bigg(x + \frac{b}{a}\bigg)^2 &= x^2 + 2\frac{b}{a}x + \bigg(\frac{b}{a}\bigg)^2\\ \end{alignedat}$

The thing we need to notice here is that on the right hand side, the constant term is the square of half the coefficient of $x$ . Thus, to complete the square,

1. Divide by the coefficient of $x^2$.
2. Identify the new coefficient of $x$, cut it in half, square it, then add it.
3. What you have can now be rewritten as a perfect square trinomial.

To understand this in full generality, let’s try it with the quadratic polynomial $ax^2 + bx$.

1. $x^2 + \frac{b}{a}x$
2. $x^2 + \frac{b}{a}x + \big( \frac{b}{2a} \big)^2$
3. $\big( x + \frac{b}{2a} \big)^2$