# Factoring 3rd degree polynomial

Can you factor (x+1) out of (x^3+1)? I would think you can, but I’m uncertain of what the other factor(s) is/are. How do you determine this?

x^3 + 1 is a sum of two cubes, so you’d use the sum of two cubes formula:

``````a^3 + b^3 = (a+b)(a^2 - a*b + b^2)
``````

With a=x and b=1, that results in (x+1)(x^2 - x + 1), which my TI-89 confirms.

Yes, you can. The factor theorem states that if you have a polynomial f(x), where f(a) = 0, then (x-a) is a factor of f(x).

In this case f(-1) = (-1)^3 + 1 =0, hence (x+1) is a factor.

By long division, you can see that (x^3+1) = (x^2 - x + 1)(x + 1)

This is also an example of the “Sum of Cubes”; (x^3 + y^3) = (x + y)(x^2 - xy + y^2)

Okay thanks. I forgot but I can also use synthetic division.

Another quick thing I decided to point out that helps.

The rational root theorem states that the only roots of a polynomial equation with integer coefficients must be a rational number where the numerator is a factor of the last coefficient and the denominator is a factor of the first coefficient.

For example, all roots of
x^3 + 1 must be in the set {± 1}.
5x^3 + 1 must be in the set{± 1, ± 1/5}.
15x^3 + 4x^2 + 4 must be in the set {±1, ±2, ±4, ±2/3, ±2/5, ±2/15, ±4/3, ±4/5, ±4/15}.

Note that this tells you only what CANNOT be a root. Numbers that satisfy this condition must still be checked for roothood* by synthetic division.
In addition, also note that “last coefficient” and “first coefficient” refer to coefficients in these positions when the polynomial is organized with the higher power of the variable first.

–Hope this random knowledge helps you one day.

*roothood - the property of being a root (fictitious word)

If you are working out of a book or off a worksheet (at least unless you are in like calculus or something) you will almost always find that x+1 or x-1 is a factor. I love my TI 89. The price is outrageous though. Do those emulator things designed for testing calculator programs have the solve and factor functions and such? If so I gotta get me one of those.