Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

closed as off-topic by Behaviour, Jack D'Aurizio, J. W. Perry, Claude Leibovici, Eric Stucky Aug 6 at 8:30

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Behaviour, Jack D'Aurizio, J. W. Perry, Claude Leibovici, Eric Stucky
If this question can be reworded to fit the rules in the help center, please edit the question.

6  
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

5 Answers 5

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

Definitly

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

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

share|improve this answer
1  
This's too complicated... –  ᴊ ᴀ s ᴏ ɴ Oct 6 '12 at 13:24
3  
@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

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

share|improve this answer

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 $x^2+2x-8=0$.

share|improve this answer

HINT:

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

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

share|improve this answer

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 $x=-4$.

share|improve this answer