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

Jul 3, 2018

The answer is $= 720$

#### Explanation:

${\left(a + b\right)}^{n} = {\sum}_{k = 0}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right) {a}^{n - k} {b}^{k}$

Where,

((n),(k))=(n!)/((n-k)!(k!))

Here, we have

${\left(2 x + 3\right)}^{5}$

And we need the coefficient of ${x}^{3}$

$n = 5$

$a = 2 x$

$b = 3$

$k = 2$

The coefficient is

$= \left(\begin{matrix}5 \\ 2\end{matrix}\right) \cdot {\left(2\right)}^{3} \cdot {\left(3\right)}^{2}$

=(5!)/((3!)(2!))* 8* 9

$= 10 \cdot 72$

$= 720$