## Algebra 1

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)