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

1 Answer
Jan 17, 2017

Answer:

#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.