Question

what is the completely factored form of d^4 - 81? (d + 3)(d - 3)(d + 3)(d - 3) (d^2 + 9)(d + 3)(d - 3) (d^2 + 9)(d - 3)(d - 3) (d^2 + 9)(d^2 - 9)

Explanation:

Step1: Recognize difference - of - squares

We know that \(a^{2}-b^{2}=(a + b)(a - b)\). Here, \(d^{4}-81=(d^{2})^{2}-9^{2}\), so \((d^{2})^{2}-9^{2}=(d^{2}+9)(d^{2}-9)\) by the difference - of - squares formula.

Step2: Factor \(d^{2}-9\) further

Since \(d^{2}-9=d^{2}-3^{2}\), using the difference - of - squares formula again, \(d^{2}-3^{2}=(d + 3)(d - 3)\). So \(d^{4}-81=(d^{2}+9)(d + 3)(d - 3)\).

Answer:

B. \((d^{2}+9)(d + 3)(d - 3)\)