Lesson Objectives

- Learn how to find all possible rational zeros.

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

In this lesson, we want to learn how to use the rational roots theorem, which is also known as the rational roots test to find all of the possible rational zeros for a polynomial function. Now, the results of this test are not necessarily zeros for our polynomial function. These results are simply candidates that can be verified using the factor theorem. Additionally, this test will only work when all coefficients are integers. If p/q is a zero of f, then p is a factor of our constant term and q is a factor of our leading coefficient. Let's look at an example.

Example #1: Find all possible rational zeros. $$f(x)=3x^3 - 7x^2 - 7x + 3$$ To build a list think about all the possible combinations of p/q that can be made.

p is a factor of the constant term, which in this case is 3. Since 3 is a prime number, 3 will only factor into 1 • 3. We also need to consider that -1 and -3 are factors as well.

q is a factor of the leading coefficient, in this case, we also have a 3. Again, 3 is a prime number, so 3 will only factor into 1 • 3. Again, we also need to consider that -1 and -3 are factors as well.

Let's just take each p and divide it by each possible q. After removing duplicates, this gives us 6 possibilities. Notice how we pull the plus or minus symbol outside to save room. This means we have 1/3, -1/3, 1, -1, 3, and -3 as candidates. $$\frac{p}{q}: \pm \left(\frac{1}{3}, 1, 3\right)$$

Example #1: Find all possible rational zeros. $$f(x)=3x^3 - 7x^2 - 7x + 3$$ To build a list think about all the possible combinations of p/q that can be made.

p is a factor of the constant term, which in this case is 3. Since 3 is a prime number, 3 will only factor into 1 • 3. We also need to consider that -1 and -3 are factors as well.

q is a factor of the leading coefficient, in this case, we also have a 3. Again, 3 is a prime number, so 3 will only factor into 1 • 3. Again, we also need to consider that -1 and -3 are factors as well.

Let's just take each p and divide it by each possible q. After removing duplicates, this gives us 6 possibilities. Notice how we pull the plus or minus symbol outside to save room. This means we have 1/3, -1/3, 1, -1, 3, and -3 as candidates. $$\frac{p}{q}: \pm \left(\frac{1}{3}, 1, 3\right)$$

#### Skills Check:

Example #1

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

Please choose the best answer.

A

$$\pm \left(1, 2, \frac{1}{2}\right)$$

B

$$\pm \left(0, 1, \frac{1}{2}\right)$$

C

$$\pm \left(7, 3, \frac{1}{2}\right)$$

D

$$\pm \left(2, 1 \right)$$

E

$$\pm \left(5, 3 \right)$$

Example #2

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

Please choose the best answer.

A

$$\pm \left(0, 1, 3 \right)$$

B

$$\pm \left(0, \frac{1}{3}, \frac{1}{27}\right)$$

C

$$\pm \left(0, 1, 3 \right)$$

D

$$\pm \left(1, 3, \frac{1}{3}\right)$$

E

$$\pm \left(1, 5, \frac{1}{3}\right)$$

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