Expert: Richard J. Raridon - 7/17/2009

Question
Hi Richard! i need help on factoring completely one each
equation

a) x^4-8x^2+16

b) 16x^4-625

c) 27x^3+64

d) x^3-y^3

yea....im not that strong when it comes to factoring
completely

Answer
a) (x^2-4)^2 = [(x+2)(x-2)]^2
b) (4x^2+25)(2x-5)(2x+5)
c) Since 27 and 64 are both cubes, one of the factors is going to be
(3x-4) or (3x+4).  Pick one and divide it into 27x^3+64.  If it divides evenly, then you will have the other factor.  In this case the factors are (3x+4) and (9x^2-12x+16)
d) here again, one factor is going to be (x-y) or (x+y). Pick one and divide.  In this case the factors are (x-y) and (x^2+xy+y^2)