Question

How do you factor completely `x^4-81`?

Answer 1

`(x^4 - 81) = (x^2+9)(x^2-9)`

`(x^2+9)(x^2-9) = (x^2+9)(x+3)(x-3)`

Answer 2

`(x-3)(x+3)(x^2+9)`

Explanation:

This is a `color(blue)"difference of squares"` and, in general, factorises as follows.

`color(red)(|bar(ul(color(white)(a/a)color(black)(a^2-b^2=(a-b)(a+b))color(white)(a/a)|)))........ (A)`

here `(x^2)^2=x^4" and " (9)^2=81`

`rArra=x^2" and " b=9`

substituting into (A)

`rArrx^4-81=(x^2-9)(x^2+9)........ (B)`

Now, the factor `x^2-9 " is also a "color(blue)"difference of squares"`

`rArrx^2-9=(x-3)(x+3)`

substituting into (B) to complete the factorising.

`rArrx^4-81=(x-3)(x+3)(x^2+9)`