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
#1:
Instructions: Find the possible rational zeros.
$$a)\hspace{.2em}f(x)=x^3 + 4x^2 + 5x + 2$$
$$b)\hspace{.2em}f(x)=x^3 + 8x^2 + 20x + 25$$
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#2:
Instructions: Find the possible rational zeros.
$$a)\hspace{.2em}f(x)=5x^3 + x^2 - 5x - 1$$
$$b)\hspace{.2em}f(x)=2x^3 + 5x^2 + 4x + 1$$
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#3:
Instructions: Find the possible rational zeros.
$$a)\hspace{.2em}f(x)=2x^3 - 10x^2 + 15x - 9$$
$$b)\hspace{.2em}f(x)=x^3 - 10x^2 - 36x - 24$$
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#4:
Instructions: Find the possible rational zeros.
$$a)\hspace{.2em}f(x)=2x^3 + 3x^2 - 1$$
$$b)\hspace{.2em}f(x)=3x^3 - 5x^2 + 7x + 3$$
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#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?
$$a)\hspace{.2em}f(x)=\frac{1}{3}x^3 - \frac{11}{6}x^2 + 3x - \frac{4}{3}$$
$$b)\hspace{.2em}f(x)=2x^3 - \frac{1}{2}x^2 - 2x + \frac{1}{2}$$
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Written Solutions:
#1:
Solutions:
$$a)\hspace{.2em}\pm 1, \pm 2$$
$$b)\hspace{.2em}\pm 1, \pm 5, \pm 25$$
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#2:
Solutions:
$$a)\hspace{.2em}\pm 1, \pm \frac{1}{5}$$
$$b)\hspace{.2em}\pm 1, \pm \frac{1}{2}$$
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#3:
Solutions:
$$a)\hspace{.2em}\pm 1, \pm 3, \pm 9, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{9}{2}$$
$$b)\hspace{.2em}\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24$$
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#4:
Solutions:
$$a)\hspace{.2em}\pm 1, \pm \frac{1}{2}$$
$$b)\hspace{.2em}\pm 1, \pm 3, \pm \frac{1}{3}$$
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#5:
Solutions:
$$a)\hspace{.2em}\pm 1, \pm 2, \pm 4, \pm 8, \pm \frac{1}{2}$$
$$b)\hspace{.2em}\pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}$$
