Find $p(0)$,$p(1)$,and $p(-2)$ for the following polynomial:
$p(y) = (y+2)(y-2)$
$p(y) = (y+2)(y-2)$
(N/A) Given the polynomial $p(y) = (y+2)(y-2)$.
Using the algebraic identity $(a+b)(a-b) = a^2 - b^2$,we can simplify the expression as $p(y) = y^2 - 4$.
Step $1$: Find $p(0)$.
$p(0) = (0)^2 - 4 = 0 - 4 = -4$.
Step $2$: Find $p(1)$.
$p(1) = (1)^2 - 4 = 1 - 4 = -3$.
Step $3$: Find $p(-2)$.
$p(-2) = (-2)^2 - 4 = 4 - 4 = 0$.
Using the algebraic identity $(a+b)(a-b) = a^2 - b^2$,we can simplify the expression as $p(y) = y^2 - 4$.
Step $1$: Find $p(0)$.
$p(0) = (0)^2 - 4 = 0 - 4 = -4$.
Step $2$: Find $p(1)$.
$p(1) = (1)^2 - 4 = 1 - 4 = -3$.
Step $3$: Find $p(-2)$.
$p(-2) = (-2)^2 - 4 = 4 - 4 = 0$.
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