# Is x^2+8x-16 a perfect square trinomial, and how do you factor it?

Jun 6, 2015

No, it's not a perfect square trinomial, because the sign of the constant term is negative.

Using the quadratic formula ${x}^{2} + 8 x - 16 = 0$ has roots

$x = \frac{- 8 \pm \sqrt{{8}^{2} - \left(4 \cdot 1 \cdot - 16\right)}}{2 \cdot 1}$

$= \frac{- 8 \pm \sqrt{128}}{2}$

$= - 4 \pm 4 \sqrt{2}$

So

${x}^{2} + 8 x - 16 = \left(x + 4 + 4 \sqrt{2}\right) \left(x + 4 - 4 \sqrt{2}\right)$

Any perfect square trinomial must be of the form:

${a}^{2} \pm 2 a b + {b}^{2} = {\left(a \pm b\right)}^{2}$