Chapter 6.5, Problem 51PPS

To determine

To find: the factor of the given polynomial and check whether the polynomial is factored or prime.

Answer to Problem 51PPS

  $x^5-16x=x\left(x^2+4\right)\left(x+2\right)\left(x-2\right)$ .

Explanation of Solution

Given:

The given polynomial is $x^5-16x$ .

Calculation:

Now, to find the factor of the given polynomial. Therefore,

  $\begin{array}{l}{x}^5-16x\left\{\text{given polynomial}\right\}\\ =x\left(x^4-16\right)\left\{\text{factor out the GCF}\right\}\end{array}$

Since are both perfect squares, so write the factor of the differences of two square.

  $\begin{array}{l}{\left(x^2\right)}^2-{\left(4\right)}^2\\ \left\{a^2-b^2=\left(a-b\right)\left(a+b\right)\right\}\\ =\left(x^2+4\right)\left(x^2-4\right)\left\{\text{simplified form}\right\}\\ =\left(x^2+4\right)\left(x+2\right)\left(x-2\right)\end{array}$

Thus, $x^5-16x=x\left(x^2+4\right)\left(x+2\right)\left(x-2\right)$

Hence, the factor of the given polynomial $x^5-16x=x\left(x^2+4\right)\left(x+2\right)\left(x-2\right)$ .