Rational Root-Rational Zero Theorem worksheet - Free Printable

Problem Overview:

The task involves using the Rational Root Theorem to determine which rational numbers are possible zeros of given polynomials. The Rational Root Theorem states that if a polynomial \( f(x) = a_n x^n + \cdots + a_1 x + a_0 \) has integer coefficients, then every rational zero of \( f(x) \) must be of the form:

\[
\frac{p}{q}
\]

where:
- \( p \) is a factor of the constant term \( a_0 \),
- \( q \) is a factor of the leading coefficient \( a_n \).

Step-by-Step Solution:

#### 1. Polynomial: \( 2x^3 - x^2 + 19x + 10 \)

- Leading coefficient (\( a_n \)): \( 2 \)
- Factors of \( 2 \): \( \pm 1, \pm 2 \)

- Constant term (\( a_0 \)): \( 10 \)
- Factors of \( 10 \): \( \pm 1, \pm 2, \pm 5, \pm 10 \)

- Possible rational zeros (\( \frac{p}{q} \)):
\[
\pm 1, \pm 2, \pm 5, \pm 10, \pm \frac{1}{2}, \pm \frac{5}{2}
\]

- Given options:
\[
\pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 10, \pm \frac{1}{2}, \pm \frac{2}{5}, \pm \frac{5}{2}, \pm \frac{5}{3}
\]

- Possible rational zeros from the options:
\[
\pm 1, \pm 2, \pm 5, \pm 10, \pm \frac{1}{2}, \pm \frac{5}{2}
\]

- NOT possible rational zeros:
\[
\pm 3, \pm 4, \pm \frac{2}{5}, \pm \frac{5}{3}
\]

#### 2. Polynomial: \( x^5 + 3x^4 - 18x^2 + 27 \)

- Leading coefficient (\( a_n \)): \( 1 \)
- Factors of \( 1 \): \( \pm 1 \)

- Constant term (\( a_0 \)): \( 27 \)
- Factors of \( 27 \): \( \pm 1, \pm 3, \pm 9, \pm 27 \)

- Possible rational zeros (\( \frac{p}{q} \)):
\[
\pm 1, \pm 3, \pm 9, \pm 27
\]

- Given options:
\[
\pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 9, \pm 27, \pm \frac{1}{3}, \pm \frac{1}{9}
\]

- Possible rational zeros from the options:
\[
\pm 1, \pm 3, \pm 9, \pm 27
\]

- NOT possible rational zeros:
\[
\pm 2, \pm 4, \pm 5, \pm 6, \pm \frac{1}{3}, \pm \frac{1}{9}
\]

#### 3. Polynomial: \( 5x^3 - 23x^2 - 65x - 22 \)

- Leading coefficient (\( a_n \)): \( 5 \)
- Factors of \( 5 \): \( \pm 1, \pm 5 \)

- Constant term (\( a_0 \)): \( -22 \)
- Factors of \( 22 \): \( \pm 1, \pm 2, \pm 11, \pm 22 \)

- Possible rational zeros (\( \frac{p}{q} \)):
\[
\pm 1, \pm 2, \pm 11, \pm 22, \pm \frac{1}{5}, \pm \frac{2}{5}, \pm \frac{11}{5}, \pm \frac{22}{5}
\]

- Given options:
\[
\pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 10, \pm 11, \pm 12, \pm 22, \pm \frac{1}{5}, \pm \frac{2}{5}, \pm \frac{11}{5}, \pm \frac{22}{5}, \pm \frac{5}{11}, \pm \frac{5}{22}
\]

- Possible rational zeros from the options:
\[
\pm 1, \pm 2, \pm 11, \pm 22, \pm \frac{1}{5}, \pm \frac{2}{5}, \pm \frac{11}{5}, \pm \frac{22}{5}
\]

- NOT possible rational zeros:
\[
\pm 3, \pm 4, \pm 5, \pm 10, \pm 12, \pm \frac{5}{11}, \pm \frac{5}{22}
\]

Final Answer:

\[
\boxed{
\begin{array}{ll}
\text{1. Possible: } \pm 1, \pm 2, \pm 5, \pm 10, \pm \frac{1}{2}, \pm \frac{5}{2} & \text{NOT: } \pm 3, \pm 4, \pm \frac{2}{5}, \pm \frac{5}{3} \\
\text{2. Possible: } \pm 1, \pm 3, \pm 9, \pm 27 & \text{NOT: } \pm 2, \pm 4, \pm 5, \pm 6, \pm \frac{1}{3}, \pm \frac{1}{9} \\
\text{3. Possible: } \pm 1, \pm 2, \pm 11, \pm 22, \pm \frac{1}{5}, \pm \frac{2}{5}, \pm \frac{11}{5}, \pm \frac{22}{5} & \text{NOT: } \pm 3, \pm 4, \pm 5, \pm 10, \pm 12, \pm \frac{5}{11}, \pm \frac{5}{22}
\end{array}
}
\]