# How do you expand (3x-2y)^5?

Jul 13, 2017

$243 {x}^{5} - 810 {x}^{4} y + 1080 {x}^{3} {y}^{2} - 720 {x}^{2} {y}^{3} + 240 x {y}^{4} - 32 {y}^{5}$

#### Explanation:

$\text{using the "color(blue)"binomial theorem}$

•color(white)(x)(x+y)^n=sum_(r=0)^n((n),(r))x^(n-r)y^r

"where" ((n),(r))=(n!)/(r!(n-r)!)

$\text{here " x=3x" and } y = - 2 y$

$\Rightarrow {\left(3 x - 2 y\right)}^{5}$

$= \left(\begin{matrix}5 \\ 0\end{matrix}\right) {\left(3 x\right)}^{5} {\left(- 2 y\right)}^{0} + \left(\begin{matrix}5 \\ 1\end{matrix}\right) {\left(3 x\right)}^{4} {\left(- 2 y\right)}^{1} + \left(\begin{matrix}5 \\ 2\end{matrix}\right) {\left(3 x\right)}^{3} {\left(- 2 y\right)}^{2} + \left(\begin{matrix}5 \\ 3\end{matrix}\right) {\left(3 x\right)}^{2} {\left(- 2 y\right)}^{3} + \left(\begin{matrix}5 \\ 4\end{matrix}\right) {\left(3 x\right)}^{1} {\left(- 2 y\right)}^{4} + \left(\begin{matrix}5 \\ 5\end{matrix}\right) {\left(3 x\right)}^{0} {\left(- 2 y\right)}^{5}$

$\text{we can obtain the binomial coefficients using the appropriate}$
$\text{row of "color(blue)"Pascal's triangle}$

$\text{for n = 5 the row of coefficients is}$

$1 \textcolor{w h i t e}{x} 5 \textcolor{w h i t e}{x} 10 \textcolor{w h i t e}{x} 10 \textcolor{w h i t e}{x} 5 \textcolor{w h i t e}{x} 1$

$= 1.243 {x}^{5} + 5.81 {x}^{4} \left(- 2 y\right) + 10.27 {x}^{3.} 4 {y}^{2} + 10.9 {x}^{2} \left(- 8 {y}^{3}\right) + 5.3 x .16 {y}^{4} + 1. - 32 {y}^{5}$

$= 243 {x}^{5} - 810 {x}^{4} y + 1080 {x}^{3} {y}^{2} - 720 {x}^{2} {y}^{3} + 240 x {y}^{4} - 32 {y}^{5}$