Chapter 8.5, Problem 3DGP

To determine

To factor: The polynomial $4a{b}^2+21+12b^2+7a$

Answer to Problem 3DGP

  $\left(4b^2+7\right)\left(a+3\right)$

Explanation of Solution

Given:

  $4a{b}^2+21+12b^2+7a$

Calculation:

The given polynomial is $4a{b}^2+21+12b^2+7a$

Make groups as shown below

  $\left(4a{b}^2+21\right)+\left(12b^2+7a\right)$

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

  $\begin{array}{l}\left(4a{b}^2+7a\right)+\left(21b^2+21\right)\\ =a\left(4b^2+7\right)+3\left(4b^2+7\right)\\ =\left(4b^2+7\right)\left(a+3\right)\end{array}$

So, the factored form is $\left(4b^2+7\right)\left(a+3\right)$