Question

What is the expansion of ${\left(x-1\right)}^4$?

Answer 1

Determine the expansion of ${\left(x-1\right)}^4$.

We know, The binomial theorem states${\left(a+b\right)}^{\mathbf{n}}\mathbf{=}\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{0}}^{\mathbf{n}}{}^{\mathbf{n}}\mathit{c}_{\mathbf{k}}\mathbf{·}\left(a^{n-k}{b}^k\right)$.

Calculate the expansion of ${\left(x-1\right)}^4$:

From the binomial theorem,

$\begin{array}{rcl}{\left(x-1\right)}^4& =& \sum _{k=0}^4\frac{4!}{\left(4-k\right)!k!}{\left(x\right)}^{4-k}{\left(-1\right)}^k\\ & \Rightarrow & {\left(x-1\right)}^4=\frac{4!}{\left(4-0\right)!0!}{\left(x\right)}^{4-0}{\left(-1\right)}^0+\frac{4!}{\left(4-1\right)!1!}{\left(x\right)}^{4-1}{\left(-1\right)}^1+\frac{4!}{\left(4-2\right)!2!}{\left(x\right)}^{4-2}{\left(-1\right)}^2+\frac{4!}{\left(4-3\right)!3!}{\left(x\right)}^{4-3}{\left(-1\right)}^3+\frac{4!}{\left(4-4\right)!4!}{\left(x\right)}^{4-4}{\left(-1\right)}^4\\ & \Rightarrow & {\left(x-1\right)}^4=1{\left(x\right)}^4{\left(-1\right)}^0+4{\left(x\right)}^3{\left(-1\right)}^1+6{\left(x\right)}^2{\left(-1\right)}^2+4{\left(x\right)}^1{\left(-1\right)}^3+1{\left(x\right)}^0{\left(-1\right)}^4\\ & \Rightarrow & {\left(x-1\right)}^4=x^4-4x^3+6x^2-4x+1\end{array}$

Hence, the required expansion is$\begin{array}{rcl}{\left(x-1\right)}^4& =& x^4-4x^3+6x^2-4x+1\end{array}$.