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 ((I mean, obviously, there are several ways)) to find where it’s equal to zero, and to write it as factors. This way? Cowboy completing the square ((I believe this method is equivalent to Po-Shen Lo’s, but I talked about it here a few years before.)).

### 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