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

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Answer 1 · 2015-06-06T14:59:06

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

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

$x = \frac{- 8 \pm \sqrt{8^{2} - (4 \cdot 1 \cdot - 16)}}{2 \cdot 1}$ $x = (-8+-sqrt(8^2-(4*1*-16)))/(2*1)$

$= \frac{- 8 \pm \sqrt{128}}{2}$ $=(-8+-sqrt(128))/2$

$= - 4 \pm 4 \sqrt{2}$ $=-4 +- 4sqrt(2)$

So

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

Any perfect square trinomial must be of the form:

$a^{2} \pm 2 a b + b^{2} = {(a \pm b)}^{2}$ $a^2+-2ab+b^2 = (a+-b)^2$

Is x^2+8x-16x2+8x−16x^2+8x-16 a perfect square trinomial, and how do you factor it?