Solve for p

(p+3)2−64

−61‾‾‾√ and 61‾‾‾√

6‾√ and 8‾√

5 and −11

−5 and 11

p = 5 and p = -11

Step-by-step explanation:

Note that the given (p+3)2−64 is the difference of two squares, for which there is the formula

(a^2 - b^2) = (a - b)(a + b).

Thus, (p+3)2−64 = [(p + 3) - 8][(p + 3) + 8]

and these last two results simplify to [p - 5][p + 11].

Setting these last two factors = 0 individually yields p = 5 and p = -11.

my guess would be b and d, but i didn’t actually get out the pen and paper. lol