Determine the x values that cause the polynomial function to be (a) zero, (b) positive, and (c) negative.
-$$f ( x ) = \left( 2 x ^ { 2 } + 3 \right) ( x - 6 ) ^ { 2 } ( x + 9 ) ^ { 3 }$$
A) (a) $\{ - 9,6 \}$, (b) $( - 9 , \infty )$, (c) $( \infty , - 9 )$
B) (a) $\{ - 9,6 \}$, (b) $( - 9,6 )$ (c) $( - 9 , \infty ) \cup ( 6 , \infty )$
C) (a) $\{ - 9,6 \}$, (b) $( - 9,6 ) \cup ( 6 , \infty ) ($ c) $( x , - 9 )$
D) (a) $\{ - 9,6 \}$, (b) $( 6 , \infty )$, (c) $( \infty , 6 )$

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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.
-$$f ( x ) = \left( 2 x ^ { 2 } + 3 \right) ( x - 6 ) ^ { 2 } ( x + 9 ) ^ { 3 }$$
A) (a) $\{ - 9,6 \}$, (b) $( - 9 , \infty )$, (c) $( \infty , - 9 )$
B) (a) $\{ - 9,6 \}$, (b) $( - 9,6 )$ (c) $( - 9 , \infty ) \cup ( 6 , \infty )$
C) (a) $\{ - 9,6 \}$, (b) $( - 9,6 ) \cup ( 6 , \infty ) ($ c) $( x , - 9 )$
D) (a) $\{ - 9,6 \}$, (b) $( 6 , \infty )$, (c) $( \infty , 6 )$

Determine the X Values That Cause the Polynomial Function to Be