How do you factor #81x^4 -625#?
1 Answer
Jan 17, 2017
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.
