# Rational Root Theorem, or Rational Zero Theorem, How to Find a Polynomial's Zeros by Hand

## (solving polynomial equations by hand)

In this section we learn the rational root theorem for polynomial functions, also known as the rational zero theorem. This will allow us to list all of the potential rational roots, or zeros, of a polynomial function, which in turn provides us with a way of finding a polynomial's rational zeros by hand.

So, for instance, by the end of this section we'll know how to show that any rational roots of the polynomial function $f(x) = 3x^3 + x^2 - 20x + 12$ must be within the following list of potential rational zeros: $\left \{ \pm \frac{1}{3}, \ \pm \frac{2}{3}, \ \pm 1, \ \pm \frac{4}{3}, \ \pm 2, \ \pm 3, \ \pm 4, \ \pm 6, \ \pm 12 \right \}$

## Rational Root Theorem

If a polynomial functions has integer coefficients, $$a_n, \ a_{n-1},$$ ... $$a_1, \ a_0$$, such that $$a_n \neq 0$$ and $$a_0 \neq 0$$: $f(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_2x^2 + a_1x + a_0$ then any rational solution to the equation: $a_nx^n + a_{n-1}x^{n-1} + \dots + a_2x^2 + a_1x + a_0 = 0$ known as a rational root, or rational zero, can be written: $x = \frac{p}{q}$ where:

• $$p$$ is a factor of $$a_0$$, the constant term
• $$q$$ is a factor of $$a_n$$, the leading coefficient.

### What this means

The rational root theorem povides us with a method for listing potential solutions to polynomial equations. Given a polynomial function, with integer coefficients, such as: $f(x) = 2x^5 - 3x^3 + 5x^2 - 7x +8$ If any of its zeros, those are the solutions to $$2x^5 - 3x^3 + 5x^2 - 7x +8 = 0$$, are rational numbers then they must be of the form: $x = \frac{\text{factor of the constant term: } 8}{\text{factor of the leading coefficient: } 2}$ Since the factors of $$8$$ are: $1, \ 2, \ 4, \ 8$ and the factors of $$2$$ are: $1, \ 2$ the rational root theorem tells us that any ratioal root must be one of the numbers in the following set: $\left \{ \pm \frac{1}{2}, \ \pm 1, \ \pm 2, \ \pm 4, \ \pm 8 \right \}$ Careful this does not mean that one of these numbers has to be a root. But: if this polynomial has a root that is a rational number then it has to within the list we just wrote.

### Tutorial: Rational Root Theorem

In this tutorial, we learn the rational root theorem, also known as the rational zero theorem, which provides us with a way of listing all of the potential rational roots, zeros, of a polynomial.

## Exercise 1

List all of the possible rational zeros for each of the following polynomials:

1. $$f(x) = x^3 - 7x^2 + 7x + 15$$
2. $$f(x) = x^4 - 4x^3 - 13x^2 + 4x + 12$$
3. $$f(x) = x^5 - 3x^4 + 7x^2 + 10$$
4. $$f(x) = 2x^3 - 6x^2 + 5x - 8$$
5. $$f(x) = 6x^3 - 11x^2 - 7x + 10$$
6. $$f(x) = 3x^6 - 4x^5 + 2x^4 - 3x + 12$$
7. $$f(x) = -2x^3 + 9x^2 +6x - 5$$
8. $$f(x) = 4x^4 - 20x^3 - 60x^2 + 20x + 56$$

Note: this exercise can be downloaded as a worksheet to practice with:

## Solution Without Working

1. For $$f(x) = x^3 - 7x^2 + 7x + 15$$, the possible rational roots are: $\left \{ \pm 1, \ \pm 3, \ \pm 5, \ \pm 15 \right \}$
2. For $$f(x) = x^4 - 4x^3 - 13x^2 + 4x + 12$$, the possible rational roots are: $\left \{ \pm 1, \ , \ \pm 2, \ \pm 3, \ \pm 4, \ \pm 6, \ \pm 12 \right \}$
3. For $$f(x) = x^5 - 3x^4 + 7x^2 + 10$$, the possible rational roots are: $\left \{ \pm 1, \ \pm 2, \ \pm 5, \ \pm 10 \right \}$
4. For $$f(x) = 2x^3 - 6x^2 + 5x - 8$$, the possible rational roots are: $\left \{ \pm \frac{1}{2}, \ \pm 1, \ \pm 2, \ \pm 4, \ \pm 8 \right \}$
5. For $$f(x) = 6x^3 - 11x^2 - 7x + 10$$, the possible rational roots are: $\left \{ \pm \frac{1}{6}, \ \pm \frac{1}{3}, \ \pm \frac{1}{2}, \ \pm \frac{2}{3}, \ \pm \frac{5}{6}, \ \pm 1, \ \pm \frac{5}{3}, \ \pm 2, \ \pm \frac{5}{2}, \ \pm \frac{10}{3}, \ \pm 5, \ \pm 10 \right \}$
6. For $$f(x) = 3x^6 - 4x^5 + 2x^4 - 3x + 12$$, the possible rational roots are: $\left \{ \pm \frac{1}{3}, \ \pm \frac{2}{3}, \ \pm 1, \ \pm \frac{4}{3}, \ \pm 2, \ \pm 5, \ \pm 10 \right \}$
7. For $$f(x) = -2x^3 + 9x^2 +6x - 5$$, the possible rational roots are: $\left \{ \pm \frac{1}{2}, \ \pm 1, \ \pm \frac{5}{2}, \ \pm 5 \right \}$
8. For $$f(x) = 4x^4 - 20x^3 - 60x^2 + 20x + 56$$, the possible rational roots are: $\left \{ \pm \frac{1}{4} , \ \pm \frac{1}{2}, \ \pm 1, \ \pm 2, \ \pm 4, \ \pm 8, \ \pm 14, \ \pm 28, \ \pm 56 \right \}$

## Method: finding a polynomial's zeros using the rational root theorem

• Step 1: use the rational root theorem to list all of the polynomial's potential zeros.
• Step 2: use "trial and error" to find out if any of the rational numbers, listed in step 1, are indeed zero of the polynomial.

The following two tutorials illustrate how the rational root theorem can be used to help find a polynomial's zeros.

### Tutorial: Rational Root Theorem

In this tutorial, we find the zeros of a polynomial using the rational root theorem. The polynomial is: $f(x) = x^3 - 2x^2 - 5x + 6$

### Tutorial: Rational Root Theorem

In this tutorial, we find the zeros of a polynomial using the rational root theorem. The polynomial is: $f(x) = 2x^4 + x^3 - 13x^2 - 6x + 6$

## Exercise 2

Using the rational root theorem, find the zeros (the roots) of each of the following polynomial functions:

1. $$f(x) = x^3 - 2x^2 - 5x + 6$$
2. $$f(x) = 5x^3 + 8x^2 - 79x + 30$$
3. $$f(x) = 3x^3 + x^2 - 12x - 4$$
4. $$f(x) = x^4 - x^3 - 19x^2 - 11x + 30$$
5. $$f(x) = 2x^4 - x^3 - 9x^2 + 4x + 4$$
6. $$f(x) = 6x^4 - 5x^3 - 14x^2 - x + 2$$
7. $$f(x) = 2x^5 - 5x^4 - 24x^3 + 41x^2 + 34x - 24$$
8. $$f(x) = 10x^5-19x^4-14x^3+23x^2+4x - 4$$

Note: this exercise can be downloaded as a worksheet to practice with:

## Solution Without Working

1. For $$f(x) = x^3 - 2x^2 - 5x + 6$$, we find the following zeros: $\left \{ -2, \ 1, \ 3 \right \}$
2. For $$f(x) = 5x^3 + 8x^2 - 79x + 30$$, we find the zeros: $\left \{ -5, \ \frac{2}{5}, \ 3 \right \}$
3. For $$f(x) = 3x^3 + x^2 - 12x - 4$$, we find the following roots (zeros): $\left \{ -2, \ - \frac{1}{3}, \ 2 \right \}$
4. For $$f(x) = x^4 - x^3 - 19x^2 - 11x + 30$$, we find the roots: $\left \{ -3, \ -2, \ - 1, \ 5 \right \}$
5. For $$f(x) = 2x^4 - x^3 - 9x^2 + 4x + 4$$, we find the roots: $\left \{ -\frac{1}{2}, \ -2, \ 1, \ 2 \right \}$
6. We find $$f(x) = 6x^4 - 5x^3 - 14x^2 - x + 2$$ has the following roots: $\left \{ -1, \ -\frac{1}{2}, \ \frac{1}{3}, \ 2 \right \}$
7. We find $$f(x) = 2x^5 - 5x^4 - 24x^3 + 41x^2 + 34x - 24$$ has the following roots: $\left \{ -3, \ -1, \ \frac{1}{2}, \ 2, \ 4 \right \}$
8. We find the polynomial $$f(x) = 10x^5-19x^4-14x^3+23x^2+4x - 4$$ has the following zeros: $\left \{ -1, \ -\frac{1}{2}, \ \frac{2}{5}, \ 1, \ 2 \right \}$