## Precalculus (10th Edition)

$x^2-4\sqrt 3x^{3/2}+18x-12\sqrt 3x^{1/2}+9$
We are given the expression: $(\sqrt x-\sqrt 3)^4$ Use the Binomial Theorem to expand the expression: $(\sqrt x-\sqrt 3)^4=\sum_{k=0}^4 \binom{4}{k}(\sqrt x)^{4-k}(-\sqrt 3)^k$ $=\binom{4}{0}(\sqrt x)^4(-\sqrt 3)^0+\binom{4}{1}(\sqrt x)^3(-\sqrt 3)^1+\binom{4}{2}(\sqrt x)^2(-\sqrt 3)^2+\binom{4}{3}(\sqrt x)^1(-\sqrt 3)^3+\binom{4}{4}(\sqrt x)^0(-\sqrt 3)^4$ $=x^2+4x^{3/2}(-\sqrt 3)+6x^1(3)+4x^{1/2}(-3\sqrt 3)+9$ $=x^2-4\sqrt 3x^{3/2}+18x-12\sqrt 3x^{1/2}+9$