what is the completely factored form of $x^{4}+8x^{2}-9$?\n$(x + 1)(x - 1)(x + 3)(x + 3)$\n$(x + 1)(x…

Explanation:

Step1: Let (y = x^{2})

The expression (x^{4}+8x^{2}-9) becomes (y^{2}+8y - 9).

Step2: Factor (y^{2}+8y - 9)

We find two numbers that multiply to (-9) and add up to (8). The numbers are (9) and (- 1). So (y^{2}+8y - 9=(y + 9)(y - 1)).

Step3: Substitute (y=x^{2}) back

We get ((x^{2}+9)(x^{2}-1)).

Step4: Factor (x^{2}-1) further

Using the difference - of - squares formula (a^{2}-b^{2}=(a + b)(a - b)), where (a=x) and (b = 1), (x^{2}-1=(x + 1)(x - 1)).

So the completely factored form of (x^{4}+8x^{2}-9) is ((x + 1)(x - 1)(x^{2}+9)).

Answer:

((x + 1)(x - 1)(x^{2}+9))