Determine the x values that cause the polynomial function to be (a) zero, (b) positive, and (c) negative.
-$$f ( x ) = ( x - 7 ) ( 3 x + 5 ) ( x + 2 )$$
A) (a) $\left\{ - 2 , - \frac { 5 } { 3 } , 7 \right\}$, (b) $( \infty , - 2 ) \cup \left( - \frac { 5 } { 3 } , 7 \right)$, (c) $\left( - 2 , - \frac { 5 } { 3 } \right) \cup ( 7 , \infty )$
B) (a) $\left\{ - 2 , - \frac { 5 } { 3 } , 7 \right\}$ (b) $\left( - 2 , - \frac { 5 } { 3 } \right) \cup ( 7 , \infty )$, (c) $( \infty , - 2 ) \cup \left( - \frac { 5 } { 3 } , 7 \right)$
C) (a) $\{ - 2 , - 5,7 \rangle$, (b) $( - 2 , - 5 ) \cup ( 7 , \infty )$, (c) $( \infty , - 2 ) \cup ( - 5,7 )$
D) (a) $\left\{ - 2 , - \frac { 5 } { 3 } , 7 \right\}$ (b) $( - 2,7 )$, (c) $( * , - 2 ) \cup ( 7 , \infty )$

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The Answer of Determine the x values that cause the polynomial function to be (a) zero, (b) positive, and (c) negative.
-$$f ( x ) = ( x - 7 ) ( 3 x + 5 ) ( x + 2 )$$
A) (a) $\left\{ - 2 , - \frac { 5 } { 3 } , 7 \right\}$, (b) $( \infty , - 2 ) \cup \left( - \frac { 5 } { 3 } , 7 \right)$, (c) $\left( - 2 , - \frac { 5 } { 3 } \right) \cup ( 7 , \infty )$
B) (a) $\left\{ - 2 , - \frac { 5 } { 3 } , 7 \right\}$ (b) $\left( - 2 , - \frac { 5 } { 3 } \right) \cup ( 7 , \infty )$, (c) $( \infty , - 2 ) \cup \left( - \frac { 5 } { 3 } , 7 \right)$
C) (a) $\{ - 2 , - 5,7 \rangle$, (b) $( - 2 , - 5 ) \cup ( 7 , \infty )$, (c) $( \infty , - 2 ) \cup ( - 5,7 )$
D) (a) $\left\{ - 2 , - \frac { 5 } { 3 } , 7 \right\}$ (b) $( - 2,7 )$, (c) $( * , - 2 ) \cup ( 7 , \infty )$

Determine the X Values That Cause the Polynomial Function to Be