Ask Uncle Colin: How to factorise this?

Dear Uncle Colin,

How would you factorise $2x^2-8x-16$?

Yonder Expression Evades Hard Algebraic Work

Howdy, YEEHAW, and thanks for your message!

It turns out that your quadratic there doesn’t factorise: $b^2-4ac$ is 192, which isn’t a square. However, there is a way ¹ to find where it’s equal to zero, and to write it as factors. This way? Cowboy completing the square ².

Cowboy Completing The Square

Suppose we’re trying to solve $2x^2-8x-16=0$ (so $a=2$, $b=-8$ and $c=-16$.)

• The sum of the solutions is $-\frac{b}{a}$ - so here, the sum of the roots is 4.
• The graph of the function is symmetrical about the mean of the roots (2), so we can write the two roots as $(2-z)$ and $(2+z)$ for some $z$.
• The product of the solutions is $\frac{c}{a}=-8$, so $(2-z)(2+z)=-8$.
• Expanding out, $4-z^2=-8$, so $z^2=12$
• The solutions are $2+\sqrt{12}$ and $2-\sqrt{12}$.

You could even write the factorised expression as $2(x-(2+\sqrt{12}))(x-(2-\sqrt{12}))$, if you were so inclined.

Hope that helps!

- Uncle Colin

Footnotes:

1. I mean, obviously, there are several ways

2. I believe this method is equivalent to Po-Shen Lo’s, but I talked about it here a few years before.