Chapter 4.3, Problem 78PPS

To determine

To calculate: The factor of the polynomial $12c^{2}x+4d^{2}y-3d^{2}x-16c^{2}y$ .

Answer to Problem 78PPS

The factor of the polynomial $12c^{2}x+4d^{2}y-3d^{2}x-16c^{2}y$ is $\left(2c-d\right)\left(2c+d\right)\left(3x-4y\right)$ .

Explanation of Solution

Given information:

The polynomial $12c^{2}x+4d^{2}y-3d^{2}x-16c^{2}y$ .

Formula used:

Factorization of a polynomial is done by grouping the similar terms together and then common factors are taken out and simplified.

Calculation:

Consider the given polynomial $12c^{2}x+4d^{2}y-3d^{2}x-16c^{2}y$ .

Recall that factorization of a polynomial is done by grouping the similar terms together and then common factors are taken out and simplified.

So, $12c^{2}x+4d^{2}y-3d^{2}x-16c^{2}y$ will be factorized as,

  $\begin{array}{c}12c^{2}x+4d^{2}y-3d^{2}x-16c^{2}y=\left(12c^{2}x-16c^{2}y\right)+\left(4d^{2}y-3d^{2}x\right)\\ =4c^2\left(3x-4y\right)-d^2\left(3x-4y\right)\\ =\left(4c^2-d^2\right)\left(3x-4y\right)\\ =\left(2c-d\right)\left(2c+d\right)\left(3x-4y\right)\end{array}$ .

Thus, the factor of the polynomial $12c^{2}x+4d^{2}y-3d^{2}x-16c^{2}y$ is $\left(2c-d\right)\left(2c+d\right)\left(3x-4y\right)$ .