# How do you use the binomial theorem to expand and simplify the expression (2x-y)^5?

Dec 1, 2017

$= - {y}^{5} + 2 \left(5 \left({y}^{4} x - 4 {x}^{2} {y}^{3} + 8 {x}^{3} {y}^{2} - 8 y {x}^{4}\right) + 16 {x}^{5}\right)$

#### Explanation:

The binomial expansion of ${\left(a y + b x\right)}^{n} = {\left(a y\right)}^{n} + n {\left(a y\right)}^{n - 1} b x + \left(\begin{matrix}n \\ 2\end{matrix}\right) {\left(a y\right)}^{n - 2} {\left(b x\right)}^{2} + \cdots + \left(\begin{matrix}n \\ n - 2\end{matrix}\right) {\left(a y\right)}^{2} {\left(b x\right)}^{n - 2} + n \left(a y\right) {\left(b x\right)}^{n - 1} + {\left(b x\right)}^{n}$

So, ${\left(- y + 2 x\right)}^{5} = {\left(- y\right)}^{5} + 5 {\left(- y\right)}^{4} \left(2 x\right) + \left(\begin{matrix}5 \\ 2\end{matrix}\right) {\left(- y\right)}^{3} {\left(2 x\right)}^{2} + \left(\begin{matrix}5 \\ 3\end{matrix}\right) {\left(- y\right)}^{2} {\left(2 x\right)}^{3} + 5 \left(- y\right) {\left(2 x\right)}^{4} + {\left(2 x\right)}^{5}$
$= - {y}^{5} + 10 {y}^{4} x - 40 {x}^{2} {y}^{3} + 80 {x}^{3} {y}^{2} - 80 y {x}^{4} + 32 {x}^{5}$
$= - {y}^{5} + 10 \left({y}^{4} x - 4 {x}^{2} {y}^{3} + 8 {x}^{3} {y}^{2} - 8 y {x}^{4}\right) + 32 {x}^{5}$
$= - {y}^{5} + 10 \left({y}^{4} x - 4 {x}^{2} {y}^{3} + 8 {x}^{3} {y}^{2} - 8 y {x}^{4}\right) + 32 {x}^{5}$
$= - {y}^{5} + 2 \left(5 \left({y}^{4} x - 4 {x}^{2} {y}^{3} + 8 {x}^{3} {y}^{2} - 8 y {x}^{4}\right) + 16 {x}^{5}\right)$