Chapter 11.8, Problem 22WE

To determine

To calculate:

Simplify the specified term

Answer to Problem 22WE

The simplification of term that containing $x^2$ is $-84x^2{y}^5$ .

Explanation of Solution

Given information:

Expression:

  ${\left(2x-y\right)}^7$

Term:

  $x^2$

Formula used:

Binomial theorem:

  ${\left(a+b\right)}^n={\displaystyle {\sum }_{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right)}{a}^{\left(n-i\right)}{b}^i$

Where,

  $a=$ first term

  $b=$ second term

Calculation:

Consider expression ${\left(2x-y\right)}^7$

Apply the binomial theorem for ${\left(2x-y\right)}^7$

  ${\left(a+b\right)}^n={\displaystyle {\sum }_{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right)}{a}^{\left(n-i\right)}{b}^i$

Where,

  $a=2x,b=-y$

So, the simplification are as follows,

  $\begin{array}{l}{\left(a+b\right)}^n={\displaystyle {\sum }_{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right)}{a}^{\left(n-i\right)}{b}^i\\ ={\displaystyle {\sum }_{i=0}^7\left(\begin{array}{c}7\\ i\end{array}\right)}2x^{\left(7-i\right)}{\left(-y\right)}^i\end{array}$

The expand summation are as follows:

  $=\frac{7!}{0!\left(7-0\right)!}{\left(2x\right)}^7{\left(-y\right)}^0+\frac{7!}{1!\left(7-1\right)!}{\left(2x\right)}^6{\left(-y\right)}^1+\mathrm{......}+\frac{7!}{7!\left(7-7\right)!}{\left(2x\right)}^0{\left(-y\right)}^7$

The term that contains the $x^2$ is the sixth term,

  $\begin{array}{l}\frac{7!}{7!\left(7-7\right)!}{\left(2x\right)}^2{\left(-y\right)}^5\\ \frac{7!}{7!\left(7-7\right)!}=\frac{42}{2!}\\ =\frac{42}{2!}{\left(2x\right)}^2{\left(-y\right)}^5\\ {\left(2x\right)}^2=2^2{x}^2\\ =2^2\cdot \frac{42}{2!}{x}^2{\left(-y\right)}^5\\ {\left(-y\right)}^5=-y^5\\ =2^2\cdot \frac{42}{2!}{x}^2{\left(-y\right)}^5\end{array}$

Remove parentheses:

  $\begin{array}{l}\left(-a\right)=-a\\ =-\frac{42}{2!}\cdot 2^2{x}^2{y}^5\end{array}$

Multiply fractions:

  $\begin{array}{l}a\cdot \frac{b}{c}=\frac{a\cdot b}{c}\\ -\frac{42\cdot 2^2{x}^2{y}^5}{2!}\\ 42\cdot 2^2{x}^2{y}^5=168x^2{y}^5\\ =-\frac{168x^2{y}^5}{2!}\\ 2!=2\\ =-\frac{168x^2{y}^5}{2!}\\ \end{array}$

Divide the numbers:

  $\begin{array}{l}\frac{168}{2}\\ =84\\ =-84x^2{y}^5\end{array}$

The answer of term that containing the $x^2$ is $-84x^2{y}^5$ .