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#