### About The Rational Zeros Theorem:

The rational zeros theorem (rational roots theorem) allows us to find all of the possible rational zeros for a polynomial function with integer coefficients. We can then use the factor theorem to find the actual rational zeros. This test will not find irrational zeros or non-real complex zeros.

Test Objectives
• Demonstrate the ability to find all potential rational zeros for a polynomial function
Rational Zeros Theorem Practice Test:

#1:

Instructions: Find the possible rational zeros.

$$a)\hspace{.2em}f(x)=x^3 + 4x^2 + 5x + 2$$

$$b)\hspace{.2em}f(x)=x^3 + 8x^2 + 20x + 25$$

#2:

Instructions: Find the possible rational zeros.

$$a)\hspace{.2em}f(x)=5x^3 + x^2 - 5x - 1$$

$$b)\hspace{.2em}f(x)=2x^3 + 5x^2 + 4x + 1$$

#3:

Instructions: Find the possible rational zeros.

$$a)\hspace{.2em}f(x)=2x^3 - 10x^2 + 15x - 9$$

$$b)\hspace{.2em}f(x)=x^3 - 10x^2 - 36x - 24$$

#4:

Instructions: Find the possible rational zeros.

$$a)\hspace{.2em}f(x)=2x^3 + 3x^2 - 1$$

$$b)\hspace{.2em}f(x)=3x^3 - 5x^2 + 7x + 3$$

#5:

Instructions: Find the possible rational zeros.

Hint: How can you can you produce a new function with integer coefficients with the same rational zeros?

$$a)\hspace{.2em}f(x)=\frac{1}{3}x^3 - \frac{11}{6}x^2 + 3x - \frac{4}{3}$$

$$b)\hspace{.2em}f(x)=2x^3 - \frac{1}{2}x^2 - 2x + \frac{1}{2}$$

Written Solutions:

#1:

Solutions:

$$a)\hspace{.2em}\pm 1, \pm 2$$

$$b)\hspace{.2em}\pm 1, \pm 5, \pm 25$$

#2:

Solutions:

$$a)\hspace{.2em}\pm 1, \pm \frac{1}{5}$$

$$b)\hspace{.2em}\pm 1, \pm \frac{1}{2}$$

#3:

Solutions:

$$a)\hspace{.2em}\pm 1, \pm 3, \pm 9, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{9}{2}$$

$$b)\hspace{.2em}\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24$$

#4:

Solutions:

$$a)\hspace{.2em}\pm 1, \pm \frac{1}{2}$$

$$b)\hspace{.2em}\pm 1, \pm 3, \pm \frac{1}{3}$$

#5:

Solutions:

$$a)\hspace{.2em}\pm 1, \pm 2, \pm 4, \pm 8, \pm \frac{1}{2}$$

$$b)\hspace{.2em}\pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}$$