<s>[INST] Find all real zeros of $f(x)=2x^3+10x^2+5x-12$ 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.
 [/INST] $2x^3+10x^2+5x-12 = 0 \implies (x+4)(2x^2 + 2x - 3) = 0$
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 $f(x)=2x^3+10x^2+5x−12$:
 
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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