How do you factor the trinomial $x^{2} + 4 x - 32$ $x^2 + 4x - 32$ ?

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Answer 1 · 2015-12-30T19:57:56

You find its roots then write them in factor form, which means equaling each root to zero.

Solving using Bhaskara:

$\frac{- 4 \pm \sqrt{4^{2} - 4 (1 (- 32))}}{2}$ $(-4+-sqrt(4^2-4(1(-32))))/2$

$\frac{- 4 \pm \sqrt{16 + 128}}{2}$ $(-4+-sqrt(16+128))/2$

$\frac{- 4 \pm \sqrt{144}}{2}$ $(-4+-sqrt(144))/2$

$\frac{- 4 \pm 12}{2}$ $(-4+-12)/2$

$x_{1} = 4 \Rightarrow$ $x_1=4 =>$ Factor form: $x - 4 = 0$ $x-4=0$
$x_{2} = - 8 \Rightarrow$ $x_2=-8 =>$ Factor form: $x + 8 = 0$ $x+8=0$

Thus, $x^{2} + 4 x - 32 = (x - 4) (x + 8)$ $x^2+4x-32=color(green)((x-4)(x+8))$

How do you factor the trinomial x2+4x−32x^2 + 4x - 32 ?