Chapter 8.5, Problem 10CYU

To determine

To factor: The polynomial $4c{d}^2+3cd-20d-15$

Answer to Problem 10CYU

  $\left(4d+3\right)\left(cd-5\right)$

Explanation of Solution

Given:

  $4c{d}^2+3cd-20d-15$

Calculation:

The given polynomial is $4c{d}^2+3cd-20d-15$

Make groups as shown below

  $\left(4c{d}^2+3cd\right)+\left(-20d-15\right)$

So, factored out the GCF from each of the group and then factored out the common factor

  $\begin{array}{l}\left(4c{d}^2+3cd\right)+\left(-20d-15\right)\\ =cd\left(4d+3\right)-5\left(4d+3\right)\\ =\left(4d+3\right)\left(cd-5\right)\end{array}$

So, the factored form is $\left(4d+3\right)\left(cd-5\right)$