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

Jun 30, 2016

${\left(x - 3\right)}^{5} = {x}^{5} - 15 {x}^{4} + 90 {x}^{3} - 270 {x}^{2} + 405 x - 243.$

#### Explanation:

The Binomial Theorem Expansion $: {\left(x + y\right)}^{n} = {x}^{n} + \text{nC_1x^(n-1)y+""nC_2x^(n-2)y^2+""nC_3x^(n-3)y^3+.........+} n {C}_{r} {x}^{n - r} {y}^{r} + \ldots \ldots + {y}^{n} .$

Letting $y = - 3 , n = 5$, we get,
${\left(x - 3\right)}^{5} = {x}^{5} + \text{5C_1x^4(-3)^1+""5C_2x^3(-3)^2+""5C_3x^2(-3)^3+} 5 {C}_{4} {x}^{1} {\left(- 3\right)}^{4} + {\left(- 3\right)}^{5.}$

Here, $\text{5C_1=""5C_(5-1)=""5C_4=5, ""5C_2=} 5 {C}_{3} = \frac{5 \cdot 4}{1 \cdot 2} = 10$

So, ${\left(x - 3\right)}^{5} = {x}^{5} + 5 {x}^{4} \left(- 3\right) + 10 {x}^{3} \left(9\right) + 10 {x}^{2} \left(- 27\right) + 5 x \left(81\right) + \left(- 243\right) .$
$= {x}^{5} - 15 {x}^{4} + 90 {x}^{3} - 270 {x}^{2} + 405 x - 243.$