Answer
8(2v+1)(2v-1)
Work Step by Step
Given the polynomial
$32v^{2}$ - 8
We see that both the terms have a common factor of 8 thus we factor the 8 out.
$8(4v^{2}$ - 1)
We use the formula for the difference of squares to apply to this question.
The difference of squares formula is:
$(a-b) (a+b) = a^{2} - b^{2}$
= $8(4v^{2}$ - 1)
*** Take the square root of $4v^{2}$ which is 2v. Becuase 2v × 2v= $4v^{2}$
*** Take the square root of 1 which is 1. Becuase 1 × 1= 1
= $8((2v)^{2}−1^{2}$)
In the given formula let 2v represents a and 1 represents b.
$8((2v)^{2}−1^{2}$) = 8(2v+1)(2v-1)
