## Intermediate Algebra for College Students (7th Edition)

$126x^5$
Our aim is to find the fifth term for $(x-1)^{9}$. General formula:$(p+q)^n=\displaystyle \binom{n}{k}p^{n-k}q^k$ and $\displaystyle \binom{n}{k}=\dfrac{n!}{k!(n-k)!}; k=2$ Now, $(x-1)^{9}=\displaystyle \binom{9}{4}(x)^{9-4}(-1)^4$ This implies, $=\dfrac{9!}{4!(9-4)!}(x)^{5}(-1)^4$ $=\dfrac{9 \times 8 \times 6 \times 5!}{4 \times 3 \times 2 \times 1(5!)}(x)^{5}$ Thus, $=126x^5$