# Expansion of (x-1)^4?

Apr 8, 2018

${\left(x - 1\right)}^{4} \equiv {x}^{4} - 4 {x}^{3} + 6 {x}^{2} - 4 x + 1$

#### Explanation:

We can expand the expression using the binomial theorem:

${\left(x - 1\right)}^{4} \equiv {\sum}_{r = 0}^{4} \left(\begin{matrix}n \\ r\end{matrix}\right) {\left(x\right)}^{r} {\left(- 1\right)}^{n - r}$

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

$\text{ } = \left(1\right) \left({x}^{4}\right) \left(1\right) + \left(4\right) \left({x}^{3}\right) \left(- 1\right) + \left(6\right) \left({x}^{2}\right) \left(1\right) +$
$\text{ } \left(4\right) \left(x\right) \left(- 1\right) + \left(1\right) \left(1\right) \left(1\right)$

$\text{ } = {x}^{4} - 4 {x}^{3} + 6 {x}^{2} - 4 x + 1$

We could also you the appropriate row from Pascal's Triangle to gain the coefficients.