How do you factor #(x^5) - (x^4) + (x^3) - (x^2) - 12x + 12#?

1 Answer
Jun 16, 2015

Answer:

#f(x) = x^5-x^4+x^3-x^2-12x+12#

#=(x-1)(x^4+x^2-12)#

#=(x-1)(x^2-3)(x^2+4)#

#=(x-1)(x-sqrt(3))(x+sqrt(3))(x^2+4)#

Explanation:

The sum of the coefficients is zero, that is:

#1-1+1-1-12+12 = 0#

So #f(1) = 0# and #(x-1)# is a factor of #f(x)#

The remaining quartic factor is actually quadratic in #x^2#, so factors just as easily into two quadratic factors:

#x^4+x^2-12 = (x^2)^2+(x^2)-12#

#=(x^2-3)(x^2+4)#

The first of these can be viewed as a difference of squares:

#x^2-3 = x^2-(sqrt(3))^2 = (x-sqrt(3))(x+sqrt(3))#

...using the difference of squares identity:

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

The remaining quadratic factor has no linear factors with real coefficients, since #x^2+4 >= 4 > 0# for all #x in RR#.