Chapter 8.7, Problem 40P

Factor each expression.

$40.48w^2+48w+12$

To determine

To factor the expression $48w^2+48w+12$

Answer to Problem 40P

The factors of the given expression are $12\left(2w+1\right)\left(2w+1\right)$ .

Explanation of Solution

Given information:

Given expression $48w^2+48w+12$

  $\begin{array}{l}\text{Given}48w^2+48w+12=12\left(4w^2+4w+1\right)\\ =12\left({\left(2w\right)}^2+2\left(2w\right)\left(1\right)+{\left(1\right)}^2\right)\end{array}$

As we know that $a^2+2ab+b^2={\left(a+b\right)}^2$

Put $a=2w,b=1$ , we get $\left({\left(2w\right)}^2+2\left(2w\right)\left(1\right)+{\left(1\right)}^2\right)={\left(2w+1\right)}^2$

Multiplying $12$ on both sides, we get

  $\begin{array}{l}12\left({\left(2w\right)}^2+2\left(2w\right)\left(1\right)+{\left(1\right)}^2\right)={\left(2w+1\right)}^2\\ =12\left(2w+1\right)\left(2w+1\right)\end{array}$

Hence, the factors of the given expression are $12\left(2w+1\right)\left(2w+1\right)$ .