Chapter 8.5, Problem 53HP

To determine

To show: Two different ways to group the terms and factor the polynomial

Answer to Problem 53HP

  $\left(4z+5\right)\left(yz+6\right)$

Explanation of Solution

Given:

  $4y{z}^2+24z+5yz+30$

Calculation:

The given expression is $4y{z}^2+24z+5yz+30$

Make group as shown below

  $\left(4y{z}^2+24z\right)+\left(5yz+30\right)$

Factored out the GCF from each group

  $4z\left(yz+6\right)+5\left(yz+6\right)$

Factored out the common factor

  $\left(4z+5\right)\left(yz+6\right)$

Now, in the second way, make group of the terms as shown below

  $\left(4y{z}^2+5yz\right)+\left(24z++30\right)$

Factored out the GCF from each group

  $yz\left(4z+5\right)+6\left(4z+5\right)$

Factored out the common factor

  $\left(4z+5\right)\left(yz+6\right)$