# How do you find the coefficient of x^5 in the expansion of (2x+3)(x+1)^8?

Jul 12, 2015

Calculate the coefficients of ${x}^{5}$ and ${x}^{4}$ in ${\left(x + 1\right)}^{8}$, then multiply by $3$ and $2$ then add to get the answer $308$

#### Explanation:

${\left(x + 1\right)}^{8} = {\sum}_{n = 0}^{n = 8} \left(\begin{matrix}8 \\ n\end{matrix}\right) {x}^{n}$

The coefficient of ${x}^{4}$ in ${\left(x + 1\right)}^{8}$ is ((8),(4)) = (8!)/(4!4!) = 70

The coefficient of ${x}^{5}$ in ${\left(x + 1\right)}^{8}$ is ((8),(5)) = (8!)/(5!3!) = 56

So the coefficient of ${x}^{5}$ in $\left(2 x + 3\right) {\left(x + 1\right)}^{8}$ is

$2 \cdot 70 + 3 \cdot 56 = 140 + 168 = 308$