Chapter 12.5, Problem 19AYU

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

${\left(x-2\right)}^6$

To determine

To find: The expansion of the given binomial expression ${\left(x - 2\right)}^6$ using Binomial Theorem.

Answer to Problem 19AYU

Solution:

The expansion of the expression is $x^6 - 12x^5 + 60x^4 - 160x^3 + 240x^2 - 192x + 64.$

Explanation of Solution

Given:

Expression is given as ${\left(x - 2\right)}^6$.

Formula used:

The Binomial Theorem:

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

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

Here $a = - 2$ and $n = 6$. Applying Binomial theorem,

$\begin{array}{l}{\left(x - 2\right)}^6 = \left(\begin{array}{c}6\\ 0\end{array}\right)x^6 + \left(\begin{array}{c}6\\ 1\end{array}\right)\left(- 2\right)x^5 + \left(\begin{array}{c}6\\ 2\end{array}\right){\left(- 2\right)}^2{x}^4 + \left(\begin{array}{c}6\\ 3\end{array}\right){\left(- 2\right)}^3{x}^3 + \left(\begin{array}{c}6\\ 4\end{array}\right){\left(- 2\right)}^4{x}^2\\ + \left(\begin{array}{c}6\\ 5\end{array}\right){\left(- 2\right)}^{5}x + \left(\begin{array}{c}6\\ 6\end{array}\right){\left(- 2\right)}^6\\ = 1.x^6 + 6.\left(- 2\right)x^5 + 15.\left(4\right)x^4 + 20.\left(- 8\right)x^3 + 15.\left(16\right)x^2 + 6.\left(- 32\right)x + 64\\ {\left(x - 2\right)}^6 = x^6 - 12x^5 + 60x^4 - 160x^3 + 240x^2 - 192x + 64\end{array}$