# Methods for Factoring Cubics

I am looking for some advice and tips/help about something. I am in calculus now and have been doing well but I recently realized to a bit of my own embarrassment that I am still not fully comfortable factoring cubics. I can do it usually, it just takes me a while and consists of mostly guess and checking, and not a certain process.

Ill give an example,

suppose I want to factor to solve for the roots $$-x^3+9x^2-15x+2=0$$

Now; I know there are a few general methods.

I know one way is to factor by grouping, but this cannot be done in the example here. I know if possible you can also factor in such a manner that you have one root and can use the quadratic formula on the other.

I think I am looking for other possibilities, mainly a nice explanation of what I believe is called the "Rational Root Theorem".

It involves something along the lines of first looking at the constant term and finding at least one root by plugging in factors of that number, then using this root and the coefficient of the highest x, you use synthetic division to find the other roots.

Anyways, I am basically looking for any explanations on methods, and if someone could show me how to solve this using synthetic division that would be great.

Update: And I am also wondering for the rational root test, if the coefficient on the $x^3$ is not a 1 or -1, would not the amount of possible factors potentially be very large? how do you deal with this?

Update 2: I solved the above question as follows; I noted x=2 was a factor and did synthetic division as

x+2 | -1 9 -15 2

| -1 7 -1 0


then did the quadratic on $-x^2+7x-1=0$

Thank you all,

• A comment on terminology: the "number with no $x$" is the constant term. – Théophile Mar 24 '15 at 21:40
• @Théophile Yes I fixed that, thanks – Quality Mar 24 '15 at 22:23
• It depends where the problem comes from. In a class setting, you are guaranteed that there will be at least one rational root-find it. Now you have a quadratic, which you know how to solve. In a more general setting, you can't count on rational roots, but usually a numerical answer is good enough, so plot the graph, find an approximation, and use a root finder. If both fail, Cardano's solution will give you an exact expression. – Ross Millikan Mar 25 '15 at 2:57

You should look up the Rational Root Theorem. Basically it says that any rational root of a polynomial with integer coefficients is of the form $\frac{a}{b}$ (in lowest terms) where $a$ divides the constant term and $b$ divides the coefficient of the highest-order term.

It's also sometimes possible to find small integer roots just "by inspection". For example, $x=1$ and $x=-1$ are easy to test just by looking at the coefficients (though neither is a root of the cubic you gave).

• Yea I guess I was also looking for more of an idea if rational root with always work – Quality Mar 24 '15 at 23:23

In a first step set $$x = y + 3$$ and you will get something like this: $${y^3} - 12y - 11 = 0$$ Then set: $$\begin{gathered} y = u + v \hfill \\ u \cdot v = 4 \hfill \\ \end{gathered}$$ should be:

$${u^3} + {v^3} - 11 + (3 \cdot u \cdot v - 12) \cdot (u + v) = 0$$

Now follows: $$\begin{gathered} {u^3} + {v^3} = 11 \hfill \\ {u^3} \cdot {v^3} = {4^3} \hfill \\ {u^3}{u^3} + {u^3}{v^3} = 11{u^3} \hfill \\ {u^6} - 11{u^3} + 64 = 0 \hfill \\ {\left( {{u^3} - \frac{{11}}{2}} \right)^2} - {\left( {\frac{{\sqrt {135} }}{2}i} \right)^2} = 0 \hfill \\ \left( {{u^3} - \frac{1}{2}\left( {11 + \sqrt {135}i } \right)} \right) \cdot \left( {{u^3} - \frac{1}{2}\left( {11 - \sqrt {135}i } \right)} \right) = 0 \hfill \\ {u^3} = \frac{1}{2}\left( {11 + \sqrt {135}i } \right) \hfill \\ {u^3} \cdot {v^3} = 64 \hfill \\ \frac{1}{2}\left( {11 + \sqrt {135}i } \right){v^3} = {u^3}{v^3} = 64 \hfill \\ {v^3} = - \frac{{128}}{{11 + \sqrt {135}i }} = \frac{1}{2}\left( {11 - \sqrt {135}i } \right) \hfill \\ \end{gathered}$$ Right combinations of $u$ and $v$ and re-substitutions will give factorization. In this case over field $\mathbb{C}$ Key I used: Cardano-Formula.

• @LearningMath: Finished. – Frieder Mar 24 '15 at 23:18
• thanks, I really appreciate the effort. However, would this be practical for me to learn/use on an exam or assignment? And can the formula be used over $\mathbb R$ ? – Quality Mar 24 '15 at 23:23
• @LearningMath: The Cardano-Formula is more of theoretical interest. It's implemented in Mathematica also. Finding roots in 3rd order polymomials are most calculated numerically. It's faster. If 3rd order polynom has "nice" coefficients, one can very well find closed expressions for its roots. But they also are not very handy to calculate with. Thank you! – Frieder Mar 25 '15 at 0:20
• @Frieder The $\sqrt{135}i$ is a manifestation that not all the roots of the cubic are real (The other two are imaginary.) – Fan Zheng Mar 25 '15 at 2:23
• @FanZheng: Yes your right. Cubic polynomials must have one real root. The other two are complex conjugate to each other. Thank's. – Frieder Mar 25 '15 at 2:46