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Please help me for solving this equation $x^3-12x+16=0$

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What have you tried? Hint: Using factor theorem, find a zero of given polynomial. Note that any rational root must be a factor of $16$ . –  TheJoker Oct 6 '12 at 13:21
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5 Answers

up vote 3 down vote accepted

This is a cubic equation. The forme of cubic equation is $x^3+px+q=0$ where

$1)$ $p$, $q$ $\in$ R

$2)$ $D=(\frac{q}{2})^2+(\frac{p}{3})^3$

$3)$ $x=u+v=\sqrt[3]{-\frac{q}{2}+\sqrt{(\frac{q}{2})^2 +(\frac{p}{3})^3}}+\sqrt[3]{-\frac{q}{2}-\sqrt{(\frac{q}{2})^2 +(\frac{p}{3})^3}}$

For the given equation we have:

$p=-12$, $q=16$, $D=(\frac{16}{2})^2+(\frac{-12}{3})^3=8^2+(-4)^3=64-64=0$.

Because $D=0$, $u=v=\sqrt[3]{\frac{-q}{2}}$ $\Rightarrow$ $u=v=\sqrt[3]{\frac{-16}{2}}=\sqrt[3]{-8}$ $\Rightarrow$ $u_1=v_1=-2$.


$x_1=2u_1$ $\Rightarrow$ $x_1=-4$

$x_2=x_3=-u_1$ $\Rightarrow$ $x_2=x_3=2$

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This's too complicated... –  ᴊ ᴀ s ᴏ ɴ Oct 6 '12 at 13:24
@jasoncube Maybe, but is gives you the general way to solve these type of cubics, even when you can't use the factor theorem of find roots by inspection. I think it is worth learning. –  Pedro Tamaroff Oct 6 '12 at 13:32
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HINT: Since $$(s-t)^3+3st(s-t)+s^3-t^3=0$$ Thus, if you could find $s,t$ such that $s^3-t^3=16$ and $3st=-12$ then $x=s-t$ is a root.

EDIT: Taking $3st=-12\implies t=-\frac{4}{s}$ and putting it int equation $s^3-t^3=16\implies s^3-(\frac{-64}{s^3})=16$ Now let $y=s^3$ which converts it into quadratic equation in $y$ and you can solve it and then $s=y^{1/3}$. after getting $s$, compute corresponding $t$ and then $x=s-t$ is a root of the given equation.

In this particular equation $x=2$ satisfies the equation. Now to find other two roots divide $x^3-12x+16$ by $x-2$ to get a quadratic equation and solve it to get other roots

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Hint: $x=2$ is a solution of the polynomial.

So $x^3-12x+16$ is divisible by $(x-2)$.

And $x^3-12x+16= (x-2)(x^2+2x-8)$.

The roots of $x^3-12x+16=0$ are $x=2$ and the roots of $x^2+2x-8=0$.

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$$ \begin{align*} x^3-12x-16&=x\underbrace{(x^2-4)}_{(x-2)(x+2)}-8x-16 \tag{1}\\ &=x(x-2)(x+2)-\underbrace{8x-16}_{-8(x+2)}\\ &=(x+2) \underbrace{(x^2-2x-8)}_{\text{quadratic equation}}=0 \\ \end{align*}$$

Now just solve the quadratic equation and you should get something like: $$(x+2)(x+a)(x+b)=0\tag{2}$$ and find: $\,\,x=-2,x=-a \,\,\text{or}\,\,x=-b$

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Can you see by inspection that 2 is a root? If so, we can write your equation as $x^{3}-12x+16=(x-2)(ax^2+bx+c)=ax^{3}+(b-2a)x^{2}+(c-2b)x-2c$ by the factor theorem. Thus, $a=1$, $b-2a=0$, $c-2b=-12 $ and $-2c=16$
Which gives $a=1$, $b=2$ and $c=-8$.
Therefore $x^{3}-12x+16=(x-2)(x^2+2x-8)=(x-2)(x-2)(x+4)=(x-2)^{2}(x+4)$
So there is a repeated root at $x=2$ and a second at $x=-4$.

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