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Hey guys I'm having a little trouble with one problem:

Find all real zeros of $$f(x)=2x^3+10x^2+5x-12.$$

I got $x=-4,(2x^2+2x-3)$. I'm just having trouble using the quadratic formula to get the rest.

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  • $\begingroup$ Hint: en.wikipedia.org/wiki/Quadratic_equation $\endgroup$
    – Amzoti
    Nov 8 '13 at 6:10
  • $\begingroup$ $ax^2+bx+c$. In your example, a=2,b=2,c=-3. Find the Discriminant and then the 2 roots $\endgroup$ Nov 8 '13 at 6:14
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    $\begingroup$ What does that quadratic in parentheses represent? Is that $f(x)/(x+4)$? Please try to express yourself as clearly as possible, even if that takes a bit more time. $\endgroup$
    – dfeuer
    Nov 8 '13 at 6:21
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    $\begingroup$ Use Cardano's (published by him and invented by Tartaglia and his assistant Ferrari) method in my opinion. There is also a nice poem of his method,each lyric is a step of his method: $\endgroup$
    – Haha
    Nov 8 '13 at 6:52
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    $\begingroup$ Tartaglia’s Poem: When the cube and its things near Add to a new number, discrete, Determine two new numbers different By that one; this feat Will be kept as a rule Their product always equal, the same, To the cube of a third Of the number of things named. Then, generally speaking, The remaining amount Of the cube roots subtracted Will be our desired count.when a cube and its things near Add to a new number, discrete en.wikipedia.org/wiki/Cubic_function $\endgroup$
    – Haha
    Nov 8 '13 at 6:52
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$2x^3+10x^2+5x-12 = 0 \implies (x+4)(2x^2 + 2x - 3) = 0$

That's where you got to; you found one of the roots to be $-4$. The next step is simple, all you need to do is solve the quadratic equation to find the other two roots. The easiest way to do it is to plug values into the Quadratic Formula (click for video).

The solution of $ax^2+bx+c=0$ is $$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}.$$

But any monkey's uncle can plug in values to a formula and solve. I'm going to show you a method called Completing the Square:

$$2x^2 + 2x - 3 = 0$$ $$\Rightarrow 2x^2 + 2x = 3$$ $$\Rightarrow x^2 + x = \frac{3}{2}$$ $$\Rightarrow x^2 + x + \left(\frac{1}{2}\right)^2 = \frac{3 \cdot 2}{2 \cdot 2} + \left(\frac{1}{2}\right)^2$$ $$\Rightarrow \left(x + \frac{1}{2}\right)^2 = \frac{6}{4} + \frac{1}{4}$$ $$\Rightarrow x+\frac{1}{2} = \pm\sqrt{\frac{7}{4}}$$ $$\Rightarrow x = \frac{1}{2}(\pm\sqrt 7 - 1)$$ So, the remaining roots are approximately $-1.82287565553$ and $0.82287565553$.

Roots of $f(x)=2x^3+10x^2+5x−12$: Graph of roots

The roots of this cubic are $-4, \frac{1}{2}(\sqrt 7 - 1), \frac{1}{2}(- \sqrt 7 - 1)$


Since your working with cubic equations, you might find it helpful to know that one can find the roots of a cubic equations by using the Cubic Formula! A direct, uncomplicated yet unpopular form of it is as follows:
The solution of $ax^3+bx^2+cx+d=0$ is

$$x = \sqrt[3]{\left({-b^3\over 27a^3}+{bc\over 6a^2}-{d\over 2a}\right)+ \sqrt{\left({-b^3 \over 27a^3}+{bc\over 6a^2}-{d\over 2a}\right)^2 +\left({c\over 3a}-{b^2 \over 9a^2}\right)^3}} \\ +\sqrt[3]{\left({-b^3\over 27a^3}+{bc\over 6a^2}-{d\over 2a}\right)- \sqrt{\left({-b^3 \over 27a^3}+{bc\over 6a^2}-{d\over 2a}\right)^2 +\left({c\over 3a}-{b^2 \over 9a^2}\right)^3}} -{b\over 3a} $$

This formula was by Gerolamo Cardano, and isn't totally complete. Infact no one recommends using it as it has certain problems with 3 real roots. But it's fun to see it this way. Better ways are listed on the wikipedia page for Cubic Functions.
But as a rule of thumb, the best method has to be "Factor, Factor, Factor"

At times it may also be useful to know a bit more about the roots of a cubic.
Let $\mathcal A$, $\mathcal B$ and $\mathcal C$ be the roots of a basic cubic function. $$\Rightarrow\mathcal{ax^3 + bx^2 + cx + d = a(x-A)(x-B)(x-C)} $$ $$\Rightarrow\mathcal{ax^3 + bx^2 + cx + d = ax^3 - a(A+B+C)x^2 + a(AB + BC + AC)x - a(A\cdot B\cdot C)}$$

If you are to compare the coefficients of $x$ on both sides, You can reach three vital relations:

  • $\mathcal{A + B + C }= \frac{-b}{a}$
  • $\mathcal{A\cdot B\cdot C} = \frac{-d}{a}$
  • $\mathcal{AB + BC + AC} = \frac{c}{a}$

Hope this helps :D

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  • $\begingroup$ Haha, Cardan's formulas don't even all show... but it should be important to say that the two cube roots involved in the computation of the three roots must have cube-roots-of-unity coefficients, and if one has one of the cube roots as a factor, the other one must have its conjugate. $\endgroup$ Nov 8 '13 at 13:12
  • $\begingroup$ @Patrick: Thank you for pointing that out. Which method of computing the roots do you recommend? $\endgroup$
    – Nick
    Nov 8 '13 at 15:55
  • $\begingroup$ Well, avoiding Cardan's formulas is probably the best way to go in general, but in case you need to use them, you just have to be careful with that the formulas mean ; the cube roots require interpretation. $\endgroup$ Nov 8 '13 at 16:50

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