Expert: Josh - 2/21/2005

Question
How would you factor:
x^3+x^2+x-3.
So how would you factor something like that.
(I know the answer is
(x-1)(x^2+2*x+3), but I want to know how to get it.And if there are more ways than one, I would like to know them as well.)
I appreciate your time that you take out to answer these questions.You are the only expert on this site that really clarifies every point of the matter so I can figure out future problems of similar structure on my own.Thank you so much.
By the way,I was the one who asked you the questions in the past about quadratic modeling,inverse variation,etc. for my MathB regent and on the New York State Regent, I received a 98(the highest score in my class.)Thank you so much for helping me study for it.

Answer
No worries Jeff. You are welcome. Good to hear that you are doing so well.

For this type of cubic polynomial, we should consider the factors of the last coefficient.

According to the fundamental theorem of algebra, we know that there are exactly three solutions over the field of complex numbers (like a+ib, where i is referred to as the imaginary number and has a value of sqrt(-1) by definition). The problem is that a cubic polynomial like x^3+px^2+qx+r=0 may have no real solution, only one real solution or three real solutions.

Let's consider these case by case.
If x^3+px^2+qx+r=0 has no real-valued solution, then, the question would be pointless, unless you have some knowledge of how complex numbers work. Then, you would be able to find the complex roots. So, let's assume that it contains at least one real solution.

In this case, we have x^3+px^2+qx+r=(x-m)(x^2+ax+b)=0. ...[#1]
Without loss of generality, the roots of the quadratic appear in complex conjugate pairs (like x=e+if, x=e-if).
The cubic polynomial has three real-numbered solutions if and only if the quadratic x^2+ax+b=0 can be factorized into linear factors, such as (x-c)(x-d).

In any case, m*b or m*c*d must be equivalent to the coefficient r. In your question, r=-3. Its factors include {(-1,3) and (3,-1)}. So, the cubic polynomial must be divisible by one of the following linear factors.
These are (x+1),(x-3),(x-1) and (x+3).
You can use trial and error to see which one works, by substituting x=-1,3,1,-3 into the cubic polynomial to check which one gives you zero.
Then, use long division, as I've shown you last time to find the quadratic expression, x^2+ax+b as given in [#1].

There is actually an explicit formula for finding the roots of a cubic polynomial (over the complex field). Further information may be found on my geocities web page, its URL is given at the bottom of this reply.
From memory, I think I have called it Fontana's formula for cubic equations (or something similar). The derivation is there. It starts out simple, but it gets harder. No one bothers remembering the formula, because it is not worth the effort. Nonetheless, some of the algebraic techniques used to derive the formula are quite interesting. Yeah, have a read through Q0008 part (i) and part (ii) in the question archive if you are interested. Let me know if there's any problem.

Cheers.