what is the completely factored form of $p^{4}-16$?\n$(p - 2)(p - 2)(p + 2)(p + 2)$\n$(p - 2)(p - 2)(p…

what is the completely factored form of $p^{4}-16$?\n$(p - 2)(p - 2)(p + 2)(p + 2)$\n$(p - 2)(p - 2)(p - 2)(p - 2)$\n$(p - 2)(p + 2)(p^{2}+2p + 4)$\n$(p - 2)(p + 2)(p^{2}+4)$

what is the completely factored form of $p^{4}-16$?\n$(p - 2)(p - 2)(p + 2)(p + 2)$\n$(p - 2)(p - 2)(p - 2)(p - 2)$\n$(p - 2)(p + 2)(p^{2}+2p + 4)$\n$(p - 2)(p + 2)(p^{2}+4)$

Answer

Explanation:

Step1: Recognize difference - of - squares

We know that (a^{2}-b^{2}=(a - b)(a + b)). Here, (p^{4}-16=(p^{2})^{2}-4^{2}). So, ((p^{2})^{2}-4^{2}=(p^{2}-4)(p^{2}+4)).

Step2: Factor (p^{2}-4) further

Since (p^{2}-4=p^{2}-2^{2}), using the difference - of - squares formula again, we get (p^{2}-2^{2}=(p - 2)(p + 2)).

Answer:

((p - 2)(p + 2)(p^{2}+4))