Lesson Objectives

- Learn how to use the Rational Zeros Theorem

## How to Find All Possible Rational Zeros Using the Rational Zeros Theorem

In this lesson, we will learn how to use the rational zeros theorem, which is also known as the rational roots theorem to find all of the possible rational zeros for a polynomial function with integer coefficients. It is very important to understand that the rational zeros theorem is only going to give us possible rational zeros. It will not tell us whether these rational numbers are actual zeros. To determine if the rational numbers are zeros, we can use the factor theorem from the previous lesson.

Example #1: Find all possible rational zeros. $$f(x)=3x^3 + x^2 - 3x - 1$$ To build a list, let's think about all the possible combinations of p/q that can be made. We know that p is a factor of the constant term, which in this case is -1. The factors of -1 are -1 and +1 only. Additionally, we know that q is a factor of the leading coefficient, which is 3 in this case. The factors of 3 are -1, 1, -3, and 3. Let's set up our possible rational zeros.

Let's start with a p-value of -1 and go through all the q-values (-1, 1, -3, and 3): $$\frac{-1}{-1} = 1$$ $$\frac{-1}{1} = -1$$ $$\frac{-1}{-3} = \frac{1}{3}$$ $$\frac{-1}{3} = -\frac{1}{3}$$ Using the other p-value of 1 will only produce duplicates: $$\frac{1}{-1} = -1$$ $$\frac{1}{1} = 1$$ $$\frac{1}{-3} = -\frac{1}{3}$$ $$\frac{1}{3} = \frac{1}{3}$$ The quicker way to do this is to just use the plus or minus symbol ("±") in each case: $$\frac{\pm 1}{\pm 1} = \pm 1$$ $$\frac{\pm 1}{\pm 3} = \pm \frac{1}{3}$$ The possible rational zeros, p/q, are: $$\pm 1, \pm \frac{1}{3}$$ Example #2: Find all possible rational zeros. $$f(x)=5x^3 - 21x^2 - 6x + 2$$ To build a list, let's think about all the possible combinations of p/q that can be made. We know that p is a factor of the constant term, which in this case is 2. The factors of 2 are -1, 1, -2, and 2. Additionally, we know that q is a factor of the leading coefficient, which is 5 in this case. The factors of 5 are -1, 1, -5, and 5. Let's set up our possible rational zeros.

Let's start with the p-values of ±1 and go through all the q-values (±1 and ±5): $$\frac{\pm 1}{\pm 1} = \pm 1$$ $$\frac{\pm 1}{\pm 5} = \pm \frac{1}{5}$$ Now, we will use the p-values of ±2 and go through all the q-values (±1 and ±5): $$\frac{\pm 2}{\pm 1} = \pm 2$$ $$\frac{\pm 2}{\pm 5} = \pm \frac{2}{5}$$ The possible rational zeros, p/q, are: $$\pm 1, \pm \frac{1}{5}, \pm 2, \pm \frac{2}{5}$$ Example #3: Find all possible rational zeros. $$f(x)=x^3 - \frac{4}{3}x^2 - \frac{17}{3}x + 2$$ In this case, we can't immediately use the rational zeros theorem since it only applies to a polynomial function with integer coefficients.

We know that a zero occurs when f(x) = 0. $$0=x^3 - \frac{4}{3}x^2 - \frac{17}{3}x + 2$$ Multiply both sides by 3, the lcd: $$0 \cdot 3 = 3\left(x^3 - \frac{4}{3}x^2 - \frac{17}{3}x + 2\right)$$ $$0 = 3x^3 - 4x^2 - 17x + 6$$ The right side of this equation is a polynomial with integer coefficients and has the same rational zeros since multiplying both sides of an equation by the same non-zero number does not change the solution set. Let's rename this polynomial function as h(x): $$h(x) = 3x^3 - 4x^2 - 17x + 6$$ We can state that any rational zeros of f(x) will also be rational zeros of h(x). It's important to understand that while we did create a new function h(x), which is 3f(x), the rational zeros will be the same. Here, p, the constant term is 6, which has factors of -1, 1, -2, 2, -3, 3, -6, and 6. Then q, the leading coefficient is 3, which has factors of -1, 1, -3, and 3. Let's set up our possible rational zeros. $$\frac{\pm 1}{\pm 1} = \pm 1$$ $$\frac{\pm 1}{\pm 3} = \pm \frac{1}{3}$$ $$\frac{\pm 2}{\pm 1} = \pm 2$$ $$\frac{\pm 2}{\pm 3} = \pm \frac{2}{3}$$ $$\frac{\pm 3}{\pm 1} = \pm 3$$ $$\frac{\pm 3}{\pm 3} = \pm 1$$ $$\frac{\pm 6}{\pm 1} = \pm 6$$ $$\frac{\pm 6}{\pm 3} = \pm 2$$ Once we remove the duplicates, we can state the possible rational zeros, p/q, are: $$\pm 1, \pm \frac{1}{3}, \pm 2, \pm \frac{2}{3}, \pm 3, \pm 6$$

### Rational Zeros Theorem

If p/q is a rational number written in lowest terms (simplified), and if p/q is a zero of f, a polynomial function with integer coefficients, then p is a factor of the constant term and q is a factor of the leading coefficient. $$f(x) = a_nx^n + a_{n - 1}x^{n - 1} + \cdots + a_1x + a_0$$ Where there are only integer coefficients and also (a_{n}≠ 0 and a_{0}≠ 0), then every rational zero of f is of the form: $$\frac{p}{q}$$ Where p and q are integers and:- p is a factor of the constant term a
_{0} - q is a factor of the leading coefficient a
_{n}

Example #1: Find all possible rational zeros. $$f(x)=3x^3 + x^2 - 3x - 1$$ To build a list, let's think about all the possible combinations of p/q that can be made. We know that p is a factor of the constant term, which in this case is -1. The factors of -1 are -1 and +1 only. Additionally, we know that q is a factor of the leading coefficient, which is 3 in this case. The factors of 3 are -1, 1, -3, and 3. Let's set up our possible rational zeros.

Let's start with a p-value of -1 and go through all the q-values (-1, 1, -3, and 3): $$\frac{-1}{-1} = 1$$ $$\frac{-1}{1} = -1$$ $$\frac{-1}{-3} = \frac{1}{3}$$ $$\frac{-1}{3} = -\frac{1}{3}$$ Using the other p-value of 1 will only produce duplicates: $$\frac{1}{-1} = -1$$ $$\frac{1}{1} = 1$$ $$\frac{1}{-3} = -\frac{1}{3}$$ $$\frac{1}{3} = \frac{1}{3}$$ The quicker way to do this is to just use the plus or minus symbol ("±") in each case: $$\frac{\pm 1}{\pm 1} = \pm 1$$ $$\frac{\pm 1}{\pm 3} = \pm \frac{1}{3}$$ The possible rational zeros, p/q, are: $$\pm 1, \pm \frac{1}{3}$$ Example #2: Find all possible rational zeros. $$f(x)=5x^3 - 21x^2 - 6x + 2$$ To build a list, let's think about all the possible combinations of p/q that can be made. We know that p is a factor of the constant term, which in this case is 2. The factors of 2 are -1, 1, -2, and 2. Additionally, we know that q is a factor of the leading coefficient, which is 5 in this case. The factors of 5 are -1, 1, -5, and 5. Let's set up our possible rational zeros.

Let's start with the p-values of ±1 and go through all the q-values (±1 and ±5): $$\frac{\pm 1}{\pm 1} = \pm 1$$ $$\frac{\pm 1}{\pm 5} = \pm \frac{1}{5}$$ Now, we will use the p-values of ±2 and go through all the q-values (±1 and ±5): $$\frac{\pm 2}{\pm 1} = \pm 2$$ $$\frac{\pm 2}{\pm 5} = \pm \frac{2}{5}$$ The possible rational zeros, p/q, are: $$\pm 1, \pm \frac{1}{5}, \pm 2, \pm \frac{2}{5}$$ Example #3: Find all possible rational zeros. $$f(x)=x^3 - \frac{4}{3}x^2 - \frac{17}{3}x + 2$$ In this case, we can't immediately use the rational zeros theorem since it only applies to a polynomial function with integer coefficients.

We know that a zero occurs when f(x) = 0. $$0=x^3 - \frac{4}{3}x^2 - \frac{17}{3}x + 2$$ Multiply both sides by 3, the lcd: $$0 \cdot 3 = 3\left(x^3 - \frac{4}{3}x^2 - \frac{17}{3}x + 2\right)$$ $$0 = 3x^3 - 4x^2 - 17x + 6$$ The right side of this equation is a polynomial with integer coefficients and has the same rational zeros since multiplying both sides of an equation by the same non-zero number does not change the solution set. Let's rename this polynomial function as h(x): $$h(x) = 3x^3 - 4x^2 - 17x + 6$$ We can state that any rational zeros of f(x) will also be rational zeros of h(x). It's important to understand that while we did create a new function h(x), which is 3f(x), the rational zeros will be the same. Here, p, the constant term is 6, which has factors of -1, 1, -2, 2, -3, 3, -6, and 6. Then q, the leading coefficient is 3, which has factors of -1, 1, -3, and 3. Let's set up our possible rational zeros. $$\frac{\pm 1}{\pm 1} = \pm 1$$ $$\frac{\pm 1}{\pm 3} = \pm \frac{1}{3}$$ $$\frac{\pm 2}{\pm 1} = \pm 2$$ $$\frac{\pm 2}{\pm 3} = \pm \frac{2}{3}$$ $$\frac{\pm 3}{\pm 1} = \pm 3$$ $$\frac{\pm 3}{\pm 3} = \pm 1$$ $$\frac{\pm 6}{\pm 1} = \pm 6$$ $$\frac{\pm 6}{\pm 3} = \pm 2$$ Once we remove the duplicates, we can state the possible rational zeros, p/q, are: $$\pm 1, \pm \frac{1}{3}, \pm 2, \pm \frac{2}{3}, \pm 3, \pm 6$$

### Proof for the Rational Zeros Theorem

Some rules of exponents we will need are listed below. Here, the bases a and b are real numbers, and the exponents, m and n are integers. $$\left(\frac{a}{b}\right)^m = \frac{a^m}{b^m}, b ≠ 0$$ $$\frac{a^m}{a^n} = a^{m - n}, a ≠ 0$$ $$a^m \cdot a^n = a^{m + n}$$ Again, we start with a polynomial function f, with the same requirements listed above. $$f(x) = a_nx^n + a_{n - 1}x^{n - 1} + \cdots + a_1x + a_0$$ Plug in p/q for x: $$f\left(\frac{p}{q}\right) = a_n\left(\frac{p}{q}\right)^n + a_{n - 1}\left(\frac{p}{q}\right)^{n - 1} + \cdots + a_1\left(\frac{p}{q}\right) + a_0$$ Since p/q is a rational zero in lowest terms: $$f\left(\frac{p}{q}\right) = 0$$ We can use this to replace f(p/q) with 0: $$a_n\left(\frac{p}{q}\right)^n + a_{n - 1}\left(\frac{p}{q}\right)^{n - 1} + \cdots + a_1\left(\frac{p}{q}\right) + a_0 = 0$$ Rewrite using the rules of exponents: $$a_n\left(\frac{p^n}{q^n}\right) + a_{n - 1}\left(\frac{p^{n - 1}}{q^{n - 1}}\right) + \cdots + a_1\left(\frac{p}{q}\right) + a_0 = 0$$ Multiply both sides by q^{n}: $$a_n \cdot \frac{p^n}{q^n} \cdot q^n + a_{n - 1} \cdot \frac{p^{n - 1}}{q^{n - 1}} \cdot q^n + \cdots + a_1 \cdot \frac{p}{q} \cdot q^n + a_0 \cdot q^n = 0 \cdot q^n$$ Simplify: $$\require{cancel}a_n \cdot \frac{p^n}{\cancel{q^n}} \cdot \cancel{q^n} + a_{n - 1} \cdot p^{n - 1} \cdot \frac{q^n}{q^{n - 1}} + \cdots + a_1 \cdot p \cdot \frac{q^n}{q^1} + a_0 \cdot q^n = 0$$ $$a_n \cdot p^n + a_{n - 1} \cdot p^{n - 1} \cdot q^{n-(n - 1)} + \cdots + a_1 \cdot p \cdot q^{n - 1} + a_0 \cdot q^n = 0$$ $$a_np^n + a_{n - 1}p^{n - 1}q + \cdots + a_1pq^{n - 1} + a_0q^n = 0$$ Subtract a_{0}q^{n}away from both sides: $$a_np^n + a_{n - 1}p^{n - 1}q + \cdots + a_1pq^{n - 1} + a_0q^n - a_0q^n = 0 - a_0q^n$$ $$a_np^n + a_{n - 1}p^{n - 1}q + \cdots + a_1pq^{n - 1} = - a_0q^n$$ Factor out p from the left side: $$a_n \cdot p^n \cdot \frac{p}{p} + a_{n - 1} \cdot p^{n - 1} \cdot \frac{p}{p} \cdot q + \cdots + a_1 \cdot p \cdot q^{n - 1} = - a_0q^n$$ Note: the p/p, which is 1, has been added in to make the rules of exponents easier to follow. $$p\left(a_n \cdot \frac{p^n}{p^1} + a_{n - 1} \cdot \frac{p^{n - 1}}{p^1} \cdot q + \cdots + a_1 \cdot q^{n - 1}\right) = - a_0q^n$$ $$p\left(a_n p^{n - 1} + a_{n - 1} p^{n - 2} q + \cdots + a_1q^{n - 1}\right) = - a_0q^n$$ Since p is a factor of the left side, it must also be a factor of the right side. We already stated that p/q is simplified, which means that p and q have no common factors other than -1 and 1. This means that p is not a factor of q^{n}, which means it must be a factor of a_{0}. We can show that q is a factor of a_{n}in a similar way. $$a_np^n + a_{n - 1}p^{n - 1}q + \cdots + a_1pq^{n - 1} + a_0q^n = 0$$ Subtract a_{n}p^{n}away from each side: $$a_{n - 1}p^{n - 1}q + \cdots + a_1pq^{n - 1} + a_0q^n = -a_np^n$$ Factor out q from the left side: $$q\left(a_{n - 1}p^{n - 1} + \cdots + a_1p \cdot q^{n - 2} + a_0 \cdot q^{n - 1}\right) = -a_np^n$$ Since q is a factor of the left side, it must also be a factor of the right side. We already stated that p/q is simplified, which means that p and q have no common factors other than -1 and 1. This means that q is not a factor of p^{n}, which means it must be a factor of a_{n}.#### Skills Check:

Example #1

Find all possible rational zeros. $$f(x)=2x^3 + x^2 - 5x + 2$$

Please choose the best answer.

A

$$\pm 1, \pm 2, \pm \frac{1}{2}$$

B

$$\pm 0, \pm 1, \pm \frac{1}{2}$$

C

$$\pm 7, \pm 3, \pm \frac{1}{2}$$

D

$$\pm 2, \pm 1 $$

E

$$\pm 5, \pm 3 $$

Example #2

Find all possible rational zeros. $$f(x)=3x^3 - 13x^2 + 13x - 3$$

Please choose the best answer.

A

$$\pm 0, \pm 1, \pm 3$$

B

$$\pm 0, \pm \frac{1}{3}, \pm \frac{1}{27}$$

C

$$\pm 0, \pm 1, \pm 3$$

D

$$\pm 1, \pm 3, \pm \frac{1}{3}$$

E

$$\pm 1, \pm 5, \pm \frac{1}{3}$$

Example #3

Find all possible rational zeros. $$f(x)= x^3 + 2x^2 - 19x + 22$$

Please choose the best answer.

A

$$\pm 1, \pm 2, \pm \frac{1}{2}$$

B

$$\pm 1, \pm 2, \pm 3, \pm 6$$

C

$$\pm 1, \pm 3, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm 6$$

D

$$\pm 1, \pm 2, \pm 11, \pm 22$$

E

$$\pm 1, \pm 2, \pm 3, \pm \frac{1}{3}, \frac{1}{2}$$

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