Chapter 6.1, Problem 35E

Solution

To write: the coefficient of $x^4$ in the expansion of given expression.

  $A.-945$

Given information:

The given expression is:

  ${\left(x-3\right)}^7$

The given options are:

1. -945
2. -35
3. -27
4. 2835

Formula Used:

Combinations of n Objects Taken r at a Time

The number of combinations of r objects taken from a group of n distinct objects is denoted by $C{}_r$ and is given by the formula:

  $C{}_r=\frac{n!}{\left(n-r\right)!\cdot r!}$ .

  $\begin{array}{l}C{}_0=1\\ C{}_n=1\end{array}$

Theorem Used:

Binomial Theorem

For any positive integer n , the binomial expansion of ${\left(a+b\right)}^n$ is:

  ${\left(a+b\right)}^n=C{}_0{a}^n{b}^0+C{}_1{a}^{n-1}{b}^1+C{}_2{a}^{n-2}{b}^2+\cdots +C{}_n{a}^0{b}^n$

Here, each term in the expansion of ${\left(a+b\right)}^n$ has the form $C{}_r{a}^{n-r}{b}^r$ where r s an integer from 0 to n .

Explanation:

By using the binomial theorem, the given expression ${\left(x-3\right)}^7$ is written in expanded form as:

  $\begin{array}{c}{\left(x-3\right)}^7={\left(x+\left(-3\right)\right)}^7\\ =C{}_0{x}^7{\left(-3\right)}^0+C{}_1{x}^{7-1}{\left(-3\right)}^1+C{}_2{x}^{7-2}{\left(-3\right)}^2+C{}_3{x}^{7-3}{\left(-3\right)}^3+\cdots +C{}_7{x}^{7-7}{\left(-3\right)}^7\end{array}$

Here, each term in the expansion has the form $C{}_r{\left(x\right)}^{7-r}{\left(-3\right)}^r$ where r s an integer from 0 to 7.

Now, the term containing $x^4$ occurs when $r=3$ :

  $\begin{array}{c}C{}_3{\left(x\right)}^{7-3}{\left(-3\right)}^3=\left(\frac{7!}{\left(7-3\right)!3!}\right)x^4\left(-27\right)\\ =\left(\frac{7!}{4!3!}\right)x^4\left(-27\right)\\ =\left(35\right)x^4\left(-27\right)\\ =-945x^4\end{array}$

So, the coefficient of $x^4$ is $-945$ .

Thus, the correct choice is option A.