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

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Answer 1 · 2017-02-07T08:21:19

$(x-(2- 2i sqrt(7)))(x-(2+ 2i sqrt(7)))=0$

$x^2-4x+32 =0$

by using completing a square,

$(x-2)^2-(2)^2+32=0$

$(x-2)^2-4+32=0$

$(x-2)^2+28=0$

$(x-2)^2=-28 =28*(-1)$

$(x-2)^2=28i^2$ , where $i^2=-1$

$x-2=+-sqrt(28i^2)=+-sqrt(7*4*i^2)$

$x=2+- 2i sqrt(7)$

$(x-2+ 2i sqrt(7))(x-2- 2i sqrt(7))=0$ or $(x-(2- 2i sqrt(7)))(x-(2+ 2i sqrt(7)))=0$

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