Determine the x values that cause the polynomial function to be (a) zero, (b) positive, and (c) negative.
-$$f ( x ) = ( 5 x + 1 ) \left( x ^ { 2 } + 2 \right) ( x - 6 )$$
A) (a) $\left\{ - \frac { 1 } { 5 } , 6 \right\} ( b ) \left( - \frac { 1 } { 5 } , 6 \right)$ (c) $\left( * , - \frac { 1 } { 5 } \right) \cup ( 6 , \infty )$
B) (a) $\left\{ \frac { 1 } { 5 } , - 6 \right\}$ (b) $\left( \cdots , - \frac { 1 } { 5 } \right) \cup ( 6 , \infty )$, (c) $\left( - \frac { 1 } { 5 } , 6 \right)$
C) (a) $\left\{ - \frac { 1 } { 5 } , 6 \right\} ( b ) \left( * , - \frac { 1 } { 5 } \right) \cup ( 6 , \infty ) , ( c ) \left( - \frac { 1 } { 5 } , 6 \right)$
D) (a) $\left\{ - \frac { 1 } { 5 } , - 1,6 \right\}$ (b) $\left( * , - \frac { 1 } { 5 } \right) \cup ( 6 , \infty )$, (c) $\left( - \frac { 1 } { 5 } , 6 \right)$

Question

Quizplus · Accepted Answer

The polynomial is zero at $x = -\frac{1}{5}$ and $x = 6$. The term $x^2 + 2$ is always positive, so it does not affect the sign of the function. The function is positive for $x < -\frac{1}{5}$ and $x > 6$, and negative for $-\frac{1}{5} < x < 6$.

Determine the X Values That Cause the Polynomial Function to Be