Chapter 8.8, Problem 22PPE

To determine

To find:The factorization of $8p^3-32p^2+28p-112$

Answer to Problem 22PPE

  $4(2p^2+7)(p-4)$

Explanation of Solution

Given information:The expression $8p^3-32p^2+28p-112$

Calculation:

  $\begin{array}{l}8p^3-32p^2+28p-112\\ =4(2p^3-8p^2+7p-28)[\text{factoring out the GCF 4]}\\ =4\{(2p^3-8p^2)+(7p-28)\}\\ =4\{2p^2(p-4)+7(p-4)\}[\text{factoring out the GCF of each of the two terms]}\\ =4(2p^2+7)(p-4)[\text{factoring out the common term }(p-4)]\end{array}$

Thus, $8p^3-32p^2+28p-112=4(2p^2+7)(p-4)$