Answer
$x^5-5x^4+10x^2-5x+1$
Work Step by Step
We are given the expression:
$(x+1)^5$
Use the Binomial Theorem to expand the expression:
$(x-1)^5=\sum_{k=0}^5 \binom{5}{k}x^{5-k}(-1)^k$
$=\binom{5}{0}x^5(-1)^0+\binom{5}{1}x^4(-1)^1+\binom{5}{2}x^3(-1)^2+\binom{5}{3}x^2(-1)^3+\binom{5}{4}x^1(-1)^4+\binom{5}{5}(-1)^5$
$=x^5-5x^4+10x^2-5x+1$
