Chapter 6.5, Problem 56PPS

To determine

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

Answer to Problem 56PPS

  $x^2\left(x+5\right)\left(x-5\right)\left(5x+2\right)\left(4x-3\right)$ .

Explanation of Solution

Given:

The given polynomialis $20x^6-7x^5-6x^4-500x^4+175x^3+150x^2$ .

Calculation:

The given polynomial is $20x^6-7x^5-6x^4-500x^4+175x^3+150x^2$ .

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

  $\begin{array}{l}20x^6-7x^5-6x^4-500x^4+175x^3+150x^2\left\{\text{given polynomial}\right\}\\ =\left(20x^6-7x^5-6x^4\right)+\left(-500x^4+175x^3+150x^2\right)\left\{\text{group the polynomial}\right\}\\ =x^4\left(20x^2-7x-6\right)-25x^2\left(20x^2-7x+6\right)\{\text{taking }{x}^4\text{and}25x^2\text{common}}\\ =\left(x^4-25x^2\right)\left(20x^2-7x+6\right)\{\text{use distributive property of product}}\end{array}$

Again, expand the factor the required polynomial to get a GCF:

  $\begin{array}{l}\left(x^4-25x^2\right)\\ =x^2\left(x^2-25\right)\left\{\text{simplify}\right\}\\ =x^2\left(x^2-{\left(5\right)}^2\right)\left\{\text{simplify}\right\}\\ \{\text{expand in the form of }{a}^2-b=\left(a+b\right)\left(a-b\right)\}\\ =x^2\left(x+5\right)\left(x-5\right)\end{array}$

Again, expand the factor the required polynomial to get a GCF:

  $\begin{array}{l}\left(20x^2-7x+6\right)\\ =20x^2-15x+8x+6\left\{\text{express in simplified form}\right\}\\ =5x\left(4x-3\right)+2\left(4x-3\right)\left\{\text{taking }\left(4x-3\right)\text{as common}\right\}\\ \left(5x+2\right)\left(4x-3\right)\left\{\text{use distributive property of product}\right\}\end{array}$

Thus, $\left(x^4-25x^2\right)\left(20x^2-7x+6\right)=x^2\left(x+5\right)\left(x-5\right)\left(5x+2\right)\left(4x-3\right)$

Hence,the factor of the given polynomial $x^2\left(x+5\right)\left(x-5\right)\left(5x+2\right)\left(4x-3\right)$ .