# Rational Root Theorem

Reference > Mathematics > Algebra > PolynomialsThe rational root theorem is a useful tool to use in finding rational solutions (if they exist) to polynomial equations.

Rational Root Theorem: If a polynomial equation with integer coefficients has any rational roots p/q, then p is a factor of the constant term, and q is a factor of the leading coefficient.

For example, consider the following cubic equation: x^{3} + 2x^{2} - x - 2 = 0.

The leading coefficient is 1, and the constant term is -2. The factors of the constant term are 1, -1, 2 and -2, and the factors of the leading coefficient are 1 and -1. That means that if this polynomial equation has any rational roots, they must be one of the following:

1/1, 2/1, -1/1, or -2/1.

In other words, rational roots could be 1, -1, 2, or -2.

This means that (x +1), (x - 1), (x + 2), or (x - 2) could be factors.

What can we do with this information? We can pick one of these and try synthetic division to see if it'll divide evenly. Let's try the factor (x - 2). This will give us the following setup:

_2_| 1 2 -1 -2 ________________

Here we go:

_2_| 1 2 -1 -2 _____2____8___14 1 4 7 12

Uh oh...we end up with a 12 at the end, so (x - 2) doesn't work. Shall we try (x + 2)?

_-2_| 1 2 -1 -2 ____-2____0____2 1 0 -1 0

Aha! It works! And now we have x^{2} - 1 as the quotient, and this factors into (x - 1)(x -1).

Thus, x^{3} + 2x^{2} - x - 2 = 0 factors to (x + 2)(x -1)(x + 1) = 0, so x = -2, 1 or -1.

Let's try another example. Use the rational root theorem to find a rational root of 2x^{3} + 5x^{2} + 6x + 2 = 0.

The factors of the leading coefficient are 1, -1, 2, and -2. The factors of the constant term are the same. Thus, the possibilities are:

1, -1, 1/2, -1/2, 2, and -2

Now here's where the limitations of the Rational Root Theorem becomes a bit more clear. There are six possibilities to try, and there's no guarantee that any of them will work. So you could try all six of them and be no better off than you were when you started.

It turns out that the following works:

_-1/2_| 2 5 6 2 ______-1__-2__-2 2 4 4 0

Since the problem asked for a rational root, and we've found one (-1/2), we're done. If we wanted to find the other roots, we could try to factor the quotient (2x^{2} + 4x + 4), or use the quadratic formula on it.

## Questions

^{3}+3x

^{2}+5 = 0?

^{3}+2x +1 = 0?

^{3}- 6 = 0?

^{3}+ 7x

^{2}+ 12x + 10 = 0

^{3}- 8 = 0

^{3}+ x

^{2}- 4x - 3 = 0

^{3}+ 5x

^{2}- x - 5 = 0

^{3}+ 5x

^{2}+ 5x - 3.