Question

How do you factor `81x^4 -625`?

Answer 1

`81x^4-625 = (3x-5)(3x+5)(9x^2+25)`

Explanation:

The difference of squares identity can be written:

`a^2-b^2 = (a-b)(a+b)`

So we find:

`81x^4-625 = (9x^2)^2-25^2`

`color(white)(81x^4-625) = (9x^2-25)(9x^2+25)`

`color(white)(81x^4-625) = ((3x)^2-5^2)(9x^2+25)`

`color(white)(81x^4-625) = (3x-5)(3x+5)(9x^2+25)`

The remaining quadratic factor is irreducible over the reals. That is, it cannot be factorised into linear factors with real coefficients.