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How to solve $x^3 - 2x^2 -16x+16=0$ ?

I tried to factor out $x^2$ but it doesn't work..any hints?

Ok I tried to use rational root theorem and have possible answer $\pm 1, \pm 2, \pm 4, \pm 8 , \pm 16$, but my calculator gives out some decimal numbers...

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To solve the cubic, we first perform a transformation to eliminate the $x^2$ term: by letting $x = y + 2/3$, we then obtain the 'depressed' cubic $$y^3 - \frac{52}{3} y + \frac{128}{27} = 0.$$ Next, recall the triple angle cosine identity: $$4 \cos^3 \theta - 3 \cos \theta = \cos 3\theta.$$ This suggests letting $y = u \cos \theta$ so that $$\begin{align*} 0 &= u^3 \cos^3 \theta - \frac{52}{3}u \cos \theta + \frac{128}{27} \\ &= \frac{u^3}{4} \left( 4 \cos^3 \theta - \frac{208}{3u^2} \cos \theta + \frac{512}{27u^3}\right).\end{align*}$$ Thus if we chose $u = \frac{4 \sqrt{13}}{3}$, it immediately follows that $$\cos 3\theta = - \frac{8}{13\sqrt{13}}.$$ Therefore, the roots are given by $$x = \frac{2}{3} + \frac{4\sqrt{13}}{3} \cos \left( \frac{1}{3} \cos^{-1} \frac{-8}{13 \sqrt{13}} + \frac{2\pi k}{3} \right), \quad k = 0, 1, 2.$$

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  • $\begingroup$ Nice method (+1). The tags may indicate that trig has not been covered yet. $\endgroup$
    – robjohn
    Feb 14 '14 at 22:01
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If we let $\alpha$ be a root, then dividing by $x-\alpha$ leaves $$ \left[x^2+(\alpha-2)x+(\alpha^2-2\alpha-16)\right](x-\alpha)=0 $$ The roots of the quadratic are $$ \frac{2-\alpha\pm\sqrt{68+4\alpha-3\alpha^2}}{2} $$ Since $f(0)=16$ and $f(1)=-1$, there is a root between $0$ and $1$. This implies that the other two roots are real since the discriminant is positive.

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  • $\begingroup$ This doesn't give a root, but if a root is calculated by approximation, this gives the other two. $\endgroup$
    – robjohn
    Feb 14 '14 at 22:03
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Given the level the question has been set at, is it at all possible there's a typo? If the equation is changed to, say, $$x^{3}-x^{2}-16x+16=0$$ then the root $x=1$ is obvious.

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Since the equation has no integer root, you have to use numerical methods (or the complicated formula for cubic equations).

The roots are :

-3.6272

0.9414

4.6858

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  • $\begingroup$ You mean "the approximated roots are" $\endgroup$ Feb 14 '14 at 21:37
  • $\begingroup$ Of course, the roots are irrational. $\endgroup$
    – Peter
    Feb 14 '14 at 21:38

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