How do you use the Binomial Theorem to expand (y - 2)^4?

Jun 23, 2016

${\left(y - 2\right)}^{4} = {y}^{4} - 8 {y}^{3} + 24 {y}^{2} - 32 y + 16$

Explanation:

According to binomial theorem,

${\left(a + b\right)}^{n} = {a}^{n} + \frac{n}{1} {a}^{n - 1} b + \frac{n \left(n - 1\right)}{1 \cdot 2} {a}^{n - 2} {b}^{2} + \frac{n \left(n - 1\right) \left(n - 2\right)}{1 \cdot 2 \cdot 3} {a}^{n - 3} {b}^{3} + \ldots . . + {b}^{n}$

Here we will find that the coefficients, such as

$\left\{1. n , \frac{n \left(n - 1\right)}{1 \cdot 2} , \frac{n \left(n - 1\right) \left(n - 2\right)}{1 \cdot 2 \cdot 3} \ldots \ldots \ldots \ldots \ldots . . n , 1\right\}$

will be as per Pascals triangle and symmetric from left to right and right to left.

As such ${\left(a - b\right)}^{4} = {a}^{4} - \frac{4}{1} {a}^{3} b + \frac{4 \cdot 3}{1 \cdot 2} {a}^{2} {b}^{2} - \frac{4 \cdot 3 \cdot 2}{1 \cdot 2 \cdot 3} a {b}^{3} + \frac{4 \cdot 3 \cdot 2 \cdot 1}{1 \cdot 2 \cdot 3 \cdot 4} {b}^{4}$

or ${\left(a - b\right)}^{4} = {a}^{4} - 4 {a}^{3} b + 6 {a}^{2} {b}^{2} - 4 a {b}^{3} + {b}^{4}$

Hence ${\left(y - 2\right)}^{4} = {y}^{4} - 4 {y}^{3} \cdot 2 + 6 {y}^{2} \cdot {2}^{2} - 4 y \cdot {2}^{3} + {2}^{4}$

or ${\left(y - 2\right)}^{4} = {y}^{4} - 8 {y}^{3} + 24 {y}^{2} - 32 y + 16$