Question 162987
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{{{x^5-32}}} First, set the expression = 0 and find the value of x.
{{{x^5-32 = 0}}} Add 32 to both sides.
{{{x^5 = 32}}} Take the fifth root of both sides.
{{{x = 2}}} So that x = 2 is a zero of the equation, then x-2 must be a factor, so divide {{{x^5-32}}} by {{{x-2}}}.
You can do this by long division or synthetic division (I have trouble showing the process on this site), but you will get the result:
{{{(x^5-32)/(x-2) = x^4+2x^3+4x^2+8x+16}}} so the expansion of {{{x^5-32}}} is:
{{{(x-2)(x^4+2x^3+4x^2+8x+16)}}}