Question

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

Answer 1

`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`.