# The Rational Root Theorem

The Rational Root Theorem is the last of our new tools. It gives us a way to make a (sometimes lengthy) list of all the possible rational roots of polynomial. Then we can winnow down the list with Descartes's Rule of Signs then test lists with the Factor Theorem.

## The Rational Root Theorem

Suppose $p(x)$ is a polynomial with integer coefficients. Then every rational root $\frac{p}{q}$, written in lowest terms, meets two conditions:

*p*is a factor of the constant term,*a*_{0}*q*is a factor of the leading coefficient,*a*_{n}

# Example 1

Find all rational roots of the polynomial $2x^2 - 5x - 3$.

The Rational Root Theorem says that we should start by looking at the factors of the constant term (-3) and the leading coefficient (2), including negatives. So the possible values of the numerators of the rational roots are ± 1 and ±3 and the possible values of the denominator are ±1 and ±2.

To organize those values and help to make sure we don't leave a combination out, I'll put them in a table with the numerator values labeling the columns and the denominator values labeling the rows.

1 | -1 | 3 | -3 | |

1 | ||||

-1 | ||||

2 | ||||

-2 |

Now I'll fill in the middle cells by dividing the column heading's value by the row heading's value.

1 | -1 | 3 | -3 | |

1 | $\frac{1}{1} = 1$ | $\frac{-1}{1} = -1$ | $\frac{3}{1} = 3$ | $\frac{-3}{1} = -3$ |

-1 | $\frac{1}{-1} = -1$ | $\frac{-1}{-1} = 1$ | $\frac{3}{-1} = 3$ | $\frac{-3}{-1} = -3$ |

2 | $\frac{1}{2}$ | $\frac{-1}{2}$ | $\frac{3}{2}$ | $\frac{-3}{2}$ |

-2 | $\frac{1}{-2} = -2$ | $\frac{-1}{-2} = 2$ | $\frac{3}{-2}$ | $\frac{-3}{-2}$ |

Those values are our final list. If we edit out all the duplicates, we get $\pm 1, \pm 3, \pm\frac{1}{2}, \pm\frac{3}{2}$.

If we evaluate the polynomial at $x=3$, we get

2(3)^2 - 5(3) - 3 = 18 - 15 - 3 = 0

So, according to the Factor Theorem, $x - 3$ must be a factor of the original polynomial.

Now, notice that there's only one sign change in the original polynomial. That means that, according to Descartes's Rule of Signs, there's only one positive root so we can throw away all the other positive possibilities on our list and focus on the negative values. Evaluating -1/2 also gives

2(-1/2)^2 - 5(-1/2) - 3 = 1/2 + 5/2 - 3 = 0

So, again according to the Factor Theorem, $x + 1/2$ is the second factor although there's a small problem. We know that there are only going to be two factors since the original polynomial is second degree but if you multiply out

$$(x + 1/2)(x - 3)$$You don't get the original polynomial because of the fraction. To fix this, it helps to remember that if $x - a = 0$ then $b(x - a) = 0$. In other words we can multiply a factor by any number and it still gives you the same solution to the polynomial. Since we want our factor to only have integers (like the original polynomial), I'll multiply it by 2 to eliminate the 1/2. That makes the second fact $2x - 1$ and the factored version of the polynomial

$$(2x + 1)(x - 3)$$You can quickly confirm this by multiplying out the expression or by using the factoring methods we discussed in the first factoring section for factoring quadratic polynomials.