# How do you use the Binomial Theorem to expand  (3x+y)^4?

Jul 6, 2017

${\left(3 x + y\right)}^{4} = 81 {x}^{4} + 108 {x}^{3} y + 54 {x}^{2} {y}^{2} + 12 x {y}^{3} + {y}^{4}$

#### Explanation:

Recall the Binomial Theorem

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

Where:

 ( (n), (r) ) =""^nC_r = (n!)/((n-r)!r!

is the combinatorial, which are also the numbers from the $n t h$ row of Pascal's Triangle

Thus we have:

${\left(3 x + y\right)}^{4} = {\sum}_{r = 0}^{4} \setminus \left(\begin{matrix}4 \\ r\end{matrix}\right) \setminus {\left(3 x\right)}^{4 - r} \setminus {\left(y\right)}^{r}$

$\text{ } = \left(\begin{matrix}4 \\ 0\end{matrix}\right) {\left(3 x\right)}^{4} + \left(\begin{matrix}4 \\ 1\end{matrix}\right) {\left(3 x\right)}^{3} y + \left(\begin{matrix}4 \\ 2\end{matrix}\right) {\left(3 x\right)}^{2} {y}^{2} + \left(\begin{matrix}4 \\ 3\end{matrix}\right) \left(3 x\right) {y}^{3} + \left(\begin{matrix}4 \\ 4\end{matrix}\right) {y}^{4}$

$\text{ } = \left(1\right) {\left(3 x\right)}^{4} + \left(4\right) {\left(3 x\right)}^{3} y + \left(6\right) {\left(3 x\right)}^{2} {y}^{2} + \left(4\right) \left(3 x\right) {y}^{3} + \left(1\right) {y}^{4}$

$\text{ } = \left(1\right) 81 {x}^{4} + \left(4\right) 27 {x}^{3} y + \left(6\right) 9 {x}^{2} {y}^{2} + \left(4\right) 3 x {y}^{3} + \left(1\right) {y}^{4}$

$\text{ } = 81 {x}^{4} + 108 {x}^{3} y + 54 {x}^{2} {y}^{2} + 12 x {y}^{3} + {y}^{4}$