Chapter 12.6, Problem 20E

To determine

To expand: the binomial by using the binomial theorem.

Answer to Problem 20E

  $81m^4+108\sqrt{2}{m}^3+108m^2+24\sqrt{2}m+4$

Explanation of Solution

Given information:

  ${\left(3m+\sqrt{2}\right)}^4$

Calculation:

For positive integer $n$ , then the following will hold true.

  ${\left(x+y\right)}^n=x^n+n{x}^{n-1}y+\frac{n\left(n-1\right)x^{n-2}{y}^2}{\left(1\right)\left(2\right)}+\frac{n\left(n-1\right)\left(n-2\right)x^{n-3}{y}^3}{\left(1\right)\left(2\right)\left(3\right)}+\mathrm{...}+y^n$

The coefficients ${}_n{\text{C}}_k$ can be written as ${}_n{\text{C}}_k=\frac{n!}{k!\left(n-k\right)!}$ .

So put $x=3m$ , $y=\sqrt{2}$ , $n=4$ and expand the binomial as follows:

  $\begin{array}{c}{\left(3m+\sqrt{2}\right)}^4={\left(3m\right)}^4+4{\left(3m\right)}^3{\left(\sqrt{2}\right)}^1+\frac{4\times 3}{2\times 1}{\left(3m\right)}^2{\left(\sqrt{2}\right)}^2+\frac{4\times 3\times 2}{3\times 2\times 1}{\left(3m\right)}^1{\left(\sqrt{2}\right)}^3+\frac{4\times 3\times 2\times 1}{4\times 3\times 2\times 1}{\left(\sqrt{2}\right)}^4\\ =81m^4+108\sqrt{2}{m}^3+108m^2+24\sqrt{2}m+4\end{array}$

Thus the expanded form of binomial is $81m^4+108\sqrt{2}{m}^3+108m^2+24\sqrt{2}m+4$ .