Chapter 6.5, Problem 27PPS

To determine

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

Answer to Problem 27PPS

  $12a{x}^2-20c{y}^2-18b{x}^2-10a{y}^2+15b{y}^2+24c{x}^2=\left(6x^2-5y^2\right)\left(2a-3b+4c\right)$ .

Explanation of Solution

Given:

The given polynomial is $12a{x}^2-20c{y}^2-18b{x}^2-10a{y}^2+15b{y}^2+24c{x}^2$ .

Calculation:

The given polynomial is $12a{x}^2-20c{y}^2-18b{x}^2-10a{y}^2+15b{y}^2+24c{x}^2$ .

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

  $\begin{array}{l}12a{x}^2-20c{y}^2-18b{x}^2-10a{y}^2+15b{y}^2+24c{x}^2\left\{\text{given polynomial}\right\}\\ =\left(12a{x}^2-18b{x}^2+24c{x}^2\right)+\left(-20c{y}^2-10a{y}^2+15b{y}^2\right)\left\{\text{group the}\text{polynomial}\right\}\\ =6x^2\left(2a-3b+4c\right)-5y^2\left(4c+2a-3b\right)\left\{\text{factor out the GCF}\right\}\\ =\left(6x^2-5y^2\right)\left(2a-3b+4c\right)\end{array}$

Thus, $12a{x}^2-20c{y}^2-18b{x}^2-10a{y}^2+15b{y}^2+24c{x}^2=\left(6x^2-5y^2\right)\left(2a-3b+4c\right)$

Hence, the factor of the given polynomial $12a{x}^2-20c{y}^2-18b{x}^2-10a{y}^2+15b{y}^2+24c{x}^2=\left(6x^2-5y^2\right)\left(2a-3b+4c\right)$ .