Can anyone factor this?

Question

-x^3 + 12x + 16. I am trying to solve for the zeros, but it seems that I have forgotten all my neat little tricks. Not a difference of cubes, or any of the common forumlas. I'm thinking maybe some kind of trick I'm not seeing.

Answer 1

This is: \begin{align*} -x^3 + 12x + 16 &= -x^2(x-4) - 4x^2 + 12x + 16\\ &= -x^2(x-4) - 4x(x-4) - 4x + 16\\ &= (x-4)(-x^2-4x-4)\\ &= -(x-4)(x^2+4x+4)\\ &= -(x-4)(x+2)^2 \end{align*}

Or use the Rational Roots Theorem and notice $-2$ is a root, and $4$ is a root, so you can then divide by $x-4$, then divide the resulting quadratic by $x+2$, and get the factorization given above.

Answer 2

see the equation $-x^3+12x+16=0$ has a root $x=-2$. Then we apply Vanishing method/Guessing method.

the factors should be something like, $-x^3-2x^2+2x^2+4x+8x+16=0$

or, $-x^2(x+2)+2x(x+2)+8(x+2)=0$

or, $(x+2)(-x^2+2x+8)=0$

or, $(x+2)(-x(x+2)+4(x+2))=0$

or, $(x+2)^2(-x+4)=0 $

so, x=-2,-2,4

Answer 3

If worse comes to worse, if you cannot guess a root of your cubic, there is a general method for finding the roots of a cubic, e.g., in: http://en.wikipedia.org/wiki/Cubic_function see in "Roots of a Cubic". Notice there is no general method for equations of (EDIT; 5, not 4) degree 5-or-higher, by Abel's results --only guy I know whose derivative names like abelian , are not capitalized. Whether there is a solution by radicals will depend on the Galois group of the equation.