How do you factor the expression #c^4 + c^3 - 12c - 12#?

1 Answer
George C.
Apr 13, 2016

#c^4+c^3-12c-12=(c-root(3)(12))(c^2+root(3)(12)c+root(3)(144))(c+1)#

Explanation:

We can factor this quartic using grouping then the difference of cubes identity, which may be written:

#a^3-b^3 = (a-b)(a^2+ab+b^2)#

with #a = c# and #b = root(3)(12)# as follows:

#c^4+c^3-12c-12#

#=(c^4+c^3)-(12c+12)#

#=c^3(c+1)-12(c+1)#

#=(c^3-12)(c+1)#

#=(c^3-(root(3)(12))^3)(c+1)#

#=(c-root(3)(12))(c^2+root(3)(12)c+root(3)(144))(c+1)#

If we allow Complex coefficients then this can be factored further as:

#=(c-root(3)(12))(c-omega root(3)(12))(c-omega^2 root(3)(12))(c+1)#

where #omega=-1/2+sqrt(3)/2i# is the primitive Complex cube root of #1#.

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