# Roots of this third degree polynomial

I've got the following polynomial $$x^3-6x^2-2x+40$$ and I want to find its roots. The only option I see at the moment is to compute all the divisors of $40$ and their inverse, and manually check if it's result is $0$. This works, because $4$ is a zero and now we can divide the polynomial by the factor $x-4$, resulting in a second degree polynomial (which is easier to solve).

I was wondering if there's any other method/idea to manually find the roots of this polynomial?

• There is: mediacru.sh/xdemgsRCo3xK – flawr Oct 7 '14 at 14:33
• $x\in\{4,\,1-\sqrt{11},\,1+\sqrt{11}\}$ – Alice Ryhl Oct 7 '14 at 14:35
• @flawr Look at the degree of 'u' in the "complete the quartic" step...having a textbook myself deriving the general solution to a quartic, it isn't nearly so simple as that diagram shows. – DanielV Jan 23 '15 at 10:35
• @DanielV I tried it with a few examples, and it did seem to work, what do you think is wrong with this diagram? – flawr Jan 23 '15 at 12:21

Hint: This can give some information about the possible location of roots, to help eliminate what you actually have to test. (Note: everything here refers to real roots and real zeroes.)

Write your polynomial as the function $$p(x) = x^3-6x^2-2x+40$$ and note that its derivative $$p'(x) = 3x^2 -12x-2$$ has exactly two distinct real zeroes $d_0<d_1$ given by $2\pm\frac16\sqrt{42}$. So, moving from left to right, the graph of $p$ rises to the left of $d_0$, falls between $d_0$ and $d_1$, and rises again to the right of $d_1$.

Then the possibilities depend on the values of $y_j \equiv p(d_j)$:

• if $y_0<0$, $p$ has exactly one zero (it is to the right of $d_1$);
• if $y_0=0$, $p$ has exactly two zeroes ($d_0$, and another to the right of $d_1$);

otherwise $y_0>0$, and we have that

• if $y_1<0$, $p$ has three zeroes (one to the left of $d_0$, one between $d_0$ and $d_1$, and one to the right of $d_1$)
• if $y_1=0$, $p$ has two zeroes (one to the left of $d_0$, and one at $d_1$;
• otherwise $y_1>0$, and $p$ has one zero (it is to the left of $d_0$)

For polynomial of degree $3$ you can use the following procedure. Assume that you guessed the solution $x_1=4$ (indeed $4^3-6\cdot 4^2-2\cdot 4+40 = 64 -96-8+40 =0)$.

You can use Horner's method to get the polynomial $p(x)=p_2x^2+p_1x+p_0$ such that $(x-4)\cdot p(x) = x^3-6x^2-2x+40$. You want to do that because $p(x)$ will be a polynomial of degree $2$ and it is easy (see here ) to find the solutions of such polynomial. In your case you have (with Horner's method)

$$\begin{array}{|c|c|c|c|c|}\hline&1&-6&-2&40\\\hline 4&0&1\cdot 4=4&-2\cdot 4=-8&-10\cdot 4=-40\\\hline&0+1=\color{blue}{\underbrace{1}_{:=p_2}}&-6+4=\color{blue}{\underbrace{-2}_{:=p_1}}&-2-8=\color{blue}{\underbrace{-10}_{:=p_0}}&40-40=\color{red}0\\\hline\end{array}$$

So your polynomial becomes $p(x) = \color{blue}1\cdot x^2\color{blue}{-2}\cdot x\color{blue}{-10}= x^2-2x-10$. The remainder, as you can see, is $\color{red}0.$ The solutions to this polynomial are

$$x_{2,3}=\dfrac{2\pm2\sqrt{11}}{2}=1\pm11$$

The solutions are therefore $x_{1,2,3}=4,1+\sqrt{11},1-\sqrt{11}$.

Your method, which starts by employing the Rational Root Theorem and then (implicitly) Factor Theorem and polynomial long division is the most elementary method and probably the quickest. Of course, it only works in certain cases. You can save a bit of work in the division step by using synthetic division.

There is a general cubic solution, but it's very tedious, and not worth it in this case.

Approximate (iterative) methods like Newton-Raphson may also be used, and they're quicker than the general cubic solution, but they don't give exact roots. Of course, if the values you get are "suspiciously" close to integers, you can just test the integers to see if they're indeed the roots the method is converging toward and short-cut the process.