Question

What is the completely factored form of x4 + 8x2 – 9?

Answer 1

ANSWER

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

EXPLANATION

The given function is

${x}^{4} + 8 {x}^{2} - 9$

Split the middle term

${x}^{4} + 9 {x}^{2} - {x}^{2} - 9$

Factor by grouping;

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

Factor further to get:

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

Apply difference of two squares to get:

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

Answer 2

Answer:

$(x^2+9)(x-1)(x+1)$

Step-by-step explanation:

We would need to let u = x^2 and use middle term factorization first. let's do this:

if u = x^2, then x^4 + 8x^2 – 9 would be

u^2+8u-9

Middle term factorization of this is:

(u+9)(u-1)

now replacing back u = x^2:

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

Using the formula a^2 - b^2 = (a+b)(a-b), we can write (x^2 - 1) as (x+1)(x-1).

So, the final factored form is:  (x^2+9)(x-1)(x+1)