Chapter 12.5, Problem 18AYU

In Problems 17-28, expand each expression using the Binomial Theorem.

${\left(x-1\right)}^5$

To determine

To find: The expansion of the given binomial expression ${\left(x - 1\right)}^5$ using Binomial Theorem.

Answer to Problem 18AYU

Solution:

The expansion of the expression is $x^5 - 5x^4 + 10x^3 - 10x^2 + 5x - \mathrm{1\; .}$

Explanation of Solution

Given:

Expression is given as ${\left(x - 1\right)}^5$.

Formula used:

The Binomial Theorem:

Let $x$ and $a$ be real numbers. For any positive integer $n$, we have

$\begin{array}{l}{\left(x + a\right)}^n = \left(\begin{array}{c}\begin{array}{c}n\\ 0\end{array}\end{array}\right)x^n + \left(\begin{array}{c}\begin{array}{c}n\\ 1\end{array}\end{array}\right)a{x}^{n - 1} + \cdots + \left(\begin{array}{c}\begin{array}{c}n\\ j\end{array}\end{array}\right)a^j{x}^{n - j} + \cdots + \left(\begin{array}{c}\begin{array}{c}n\\ n\end{array}\end{array}\right)a^n\\ {\left(x + a\right)}^n = {\displaystyle \sum _{k = 0}^n\left(\begin{array}{c}\begin{array}{c}n\\ k\end{array}\end{array}\right)}{x}^{n - j}{a}^j\end{array}$

Here $a = - 1$ and $n = 5$. Applying Binomial theorem,

$\begin{array}{l}{\left(x - 1\right)}^5 = \left(\begin{array}{c}\begin{array}{c}5\\ 0\end{array}\end{array}\right)x^5 + \left(\begin{array}{c}\begin{array}{c}5\\ 1\end{array}\end{array}\right)\left( - 1\right)x^4 + \left(\begin{array}{c}\begin{array}{c}5\\ 2\end{array}\end{array}\right){\left( - 1\right)}^2{x}^3 + \left(\begin{array}{c}\begin{array}{c}5\\ 3\end{array}\end{array}\right){\left( - 1\right)}^3{x}^2 + \left(\begin{array}{c}\begin{array}{c}5\\ 4\end{array}\end{array}\right){\left( - 1\right)}^4{x}^1 + \left(\begin{array}{c}\begin{array}{c}5\\ 5\end{array}\end{array}\right){\left( - 1\right)}^5\\ = 1.x^5 + 5.\left( - 1\right)x^4 + 10.\left(1\right)x^3 + 10.\left( - 1\right)x^2 + 5.\left(1\right)x + \left( - 1\right)1\\ {\left(x - 1\right)}^5 = x^5 - 5x^4 + 10x^3 - 10x^2 + 5x - 1\end{array}$