# How do you use the Binomial Theorem to expand (3x - 2y^2)^5?

${\left(3 x - 2 {y}^{2}\right)}^{5} =$
$243 {x}^{5} - 810 {x}^{4} {y}^{2} + 1080 {x}^{3} {y}^{4} - 720 {x}^{2} {y}^{6} + 240 x {y}^{8} - 32 {y}^{10}$

#### Explanation:

The formula of Binomial Theorem

${\left(a + b\right)}^{n} = \left(\begin{matrix}n \\ 0\end{matrix}\right) {a}^{n} {b}^{0} + \left(\begin{matrix}n \\ 1\end{matrix}\right) {a}^{n - 1} {b}^{1} + \left(\begin{matrix}n \\ 2\end{matrix}\right) {a}^{n - 2} {b}^{2} + \ldots +$
$\left(\begin{matrix}n \\ n - 1\end{matrix}\right) {a}^{1} {b}^{n - 1} + \left(\begin{matrix}n \\ n\end{matrix}\right) {a}^{0} {b}^{n}$

Let $a = 3 x$ and $b = - 2 {y}^{2}$

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

(3x)^5(-2y^2)^0+5/(1!)(3x)^4(-2y^2)^1+(5*4)/(2!)(3x)^3(-2y^2)^2+(5*4*3)/(3!)(3x)^2(-2y^2)^3+(5*4*3*2)/(4!)(3x)^1(2y^2)^4+(5*4*3*2*1)/(5!)(3x)^0(-2y^2)^5

$= 243 {x}^{5} - 5 \left(81\right) \left(2\right) {x}^{4} {y}^{2} + 10 \left(27\right) \left(4\right) {x}^{3} {y}^{4} + 10 \left(9\right) \left(- 8\right) {x}^{2} {y}^{6} + 5 \left(3\right) \left(16\right) x {y}^{8} - 32 {y}^{10}$

$= 243 {x}^{5} - 810 {x}^{4} {y}^{2} + 1080 {x}^{3} {y}^{4} - 720 {x}^{2} {y}^{6} + 240 x {y}^{8} - 32 {y}^{10}$

God bless you...