What is the answer when factoring the expression completely over the complex numbers?

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1 Answer
George C.
Nov 24, 2017

#y^3+2y^2+16y+32 = (y-4i)(y+4i)(y+2)#

Explanation:

The difference of squares identity can be written:

#A^2-B^2 = (A-B)(A+B)#

Use this with #A=y# and #B=4i# below...

Given:

#y^3+2y^2+16y+32#

First note that the ratio between the first and second terms is the same as that between the third and fourth terms.

So this quadrinomial will factor by grouping:

#y^3+2y^2+16y+32 = (y^3+2y^2)+(16y+32)#

#color(white)(y^3+2y^2+16y+32) = y^2(y+2)+16(y+2)#

#color(white)(y^3+2y^2+16y+32) = (y^2+16)(y+2)#

#color(white)(y^3+2y^2+16y+32) = (y^2+4^2)(y+2)#

#color(white)(y^3+2y^2+16y+32) = (y^2-(4i)^2)(y+2)#

#color(white)(y^3+2y^2+16y+32) = (y-4i)(y+4i)(y+2)#

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