How do I use the Binomial Theorem to expand this binomial? (x - 5y)^5

Aug 14, 2018

${x}^{5} - 25 {x}^{4} y + 250 {x}^{3} {y}^{2} - 1250 {x}^{2} {y}^{3} + 3125 x {y}^{4} - 3125 {y}^{5}$

Explanation:

The binomial theorem states that
${\left(a + b\right)}^{n} = {\sum}_{i = 0}^{n} C \left(n , i\right) {a}^{i} {b}^{n - i}$
where $C \left(n , i\right)$ is $n$ choose $i$.

Let's right out the coefficients of the $C \left(5 , i\right)$. I do this based on my memory of Pascal's triangle. 1, 5, 10, 10, 5, 1

From this, we can write out each of the terms, with $a = x$ and $b = - 5 y$ and $n = 5$:
${\left(x - 5 y\right)}^{5} = {\sum}_{i = 0}^{5} C \left(5 , i\right) {x}^{i} {\left(- 5 y\right)}^{5 - i}$
$= 1 {x}^{0} {\left(- 5 y\right)}^{5} + 5 {x}^{1} {\left(- 5 y\right)}^{4} + 10 {x}^{2} {\left(- 5 y\right)}^{3} + 10 {x}^{3} {\left(- 5 y\right)}^{2} + 5 {x}^{4} {\left(- 5 y\right)}^{1} + 1 {x}^{5} {\left(- 5 y\right)}^{0}$
$= - {5}^{5} {y}^{5} + {5}^{5} x {y}^{4} - 2 \cdot {5}^{4} {x}^{2} {y}^{3} + 2 \cdot {5}^{3} {x}^{3} {y}^{2} - {5}^{2} {x}^{4} y + {x}^{5}$
$= {x}^{5} - 25 {x}^{4} y + 250 {x}^{3} {y}^{2} - 1250 {x}^{2} {y}^{3} + 3125 x {y}^{4} - 3125 {y}^{5}$