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-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.

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It looks like $4$ is a root. A "trick" you could use is the rational roots theorem. You have only to check divisors of $16$ (positive and negative). –  NotNotLogical May 14 at 0:16
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Considering the rational root theorem, by inspection, $-2$ is a root. –  heropup May 14 at 0:17
    
Since $x=4$ is a root by dividing with $x-4$ you obtain a degree 2 polynomial which you can factor. –  Test123 May 14 at 0:17
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ahh yeah. I forgot that little trick. Thanks yall –  user149635 May 14 at 0:18

3 Answers 3

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.

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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

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+1 Interesting... what is the algorithm here? –  NotNotLogical May 14 at 0:36
    
In our High School our teacher used to call it Vanishing method. Here at first you need to guess a root by checking arbitrarily (thus applicable for simple equations). Then once you know a root the factoring is easy. Here the second step is computed at first and then the first step is computed. That is when you compute the second step the first step is vanished. That's the beauty of this method :). –  Sayan Bandyapadhyay May 14 at 0:42
    
I'm sorry I don't understand. How did you know to break the polynomial up as you did? Why $\pm 2x^2$ instead of $\pm 3x^2$ for example –  NotNotLogical May 14 at 0:49
    
Let me explain it in details. See here I know that $x=-2$ is a factor, then how can I write the equation as $(x+2)(-x^2+ax+b)=0$? The quadratic $-x^2$ (- as there is a - before $x^3$) factor should come from $-x^3$, so write $-x^2(x+2)+\ldots$. This is the second line. Thus the previous line I can guess is like $-x^3-2x^2+\ldots$. Thus to summarize $-2x^2$ as my guess was $x=-2$. –  Sayan Bandyapadhyay May 14 at 0:59

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.

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Slight point: There is no general solution for equations of degree 5 or higher. A formula for quartics does exist. –  Daniel Littlewood May 14 at 0:31
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@DanielLittlewood: Right, sorry, Abel (abel) is shaking on his grave. I just edited. –  user99680 May 14 at 0:32

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