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 way¹ to find where it’s equal to zero, and to write it as factors. This way? Cowboy completing the square².

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