Find the root for a third degree polynomial?

Question

So far in this course we have not been given any formula for solving third degrees polynomials.$$\frac{1}{3}x^3-2x^2+4x$$

I was thinking about doing it like this
$$x(\frac{1}{3}x^2-2x+4).$$ But that didn't help because the solutions to the roots of that one is in the complex plane which is a foreign word in this book.
It is a third degree so I know it can have at most three roots, there is only one ($x=0$) from looking at the graph. But I still don't get how to do this by calculation.
Is there some "trick" I don't see or something I can assume?

Answer 1

You've already shown that $x=0$ is the only real root and that the other two are not real.

Answer 2

To solve a polynomial you need an equation ($=0$) so to solve $$\frac{1}{3}x^3-2x^2+4x=0$$ your idea of $$x(\tfrac{1}{3}x^2-2x+4)=0$$ is the right way to go. This implies $x=0$ or $\frac{1}{3}x^2-2x+4=0$. You should be able to deal with this to spot the first of these gives the one real root.

Answer 3

If $f$ and $g$ are two polynomials, then $$f(a)g(a)=0\iff f(a)=0\;\;\text{ or }\;\;g(a)=0.$$ So, the roots of the product of the polynomials, $fg$, is just the roots of $f$, together with the roots of $g$.

In your case, $f=x$ and $g=\frac{1}{3}x^2-2x+4$. Obviously the only root of $f$ is 0; the roots of $g$ can be found with the quadratic formula: $$a=\frac{2\pm\sqrt{(-2)^2-4\cdot4\cdot\frac{1}{3}}}{2\cdot\frac{1}{3}}=\frac{2\pm2\sqrt{-\frac{1}{3}}}{2\cdot\frac{1}{3}}=3\pm i\sqrt{3}$$ Because your course has not gotten to complex numbers yet, these are not treated as solutions, but nevertheless you have verified to yourself via computation that there are no real roots of $$fg=\tfrac{1}{3}x^3-2x^2+4x$$ besides 0, because you have found all 3 of the complex roots: $0$, $3+i\sqrt{3}$, and $3-i\sqrt{3}$, and the only one that is a real number is 0.