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Determine the X Values That Cause the Polynomial Function to Be

Question 222

Multiple Choice

Determine the x values that cause the polynomial function to be (a) zero, (b) positive, and (c) negative.
- f(x) =(5x+1) (x2+2) (x−6) f ( x ) = ( 5 x + 1 ) \left( x ^ { 2 } + 2 \right) ( x - 6 )


A) (a) {−15,6}(b) (−15,6) \left\{ - \frac { 1 } { 5 } , 6 \right\} ( b ) \left( - \frac { 1 } { 5 } , 6 \right) (c) (∗,−15) ∪(6,∞) \left( * , - \frac { 1 } { 5 } \right) \cup ( 6 , \infty )
B) (a) {15,−6}\left\{ \frac { 1 } { 5 } , - 6 \right\} (b) (⋯ ,−15) ∪(6,∞) \left( \cdots , - \frac { 1 } { 5 } \right) \cup ( 6 , \infty ) , (c) (−15,6) \left( - \frac { 1 } { 5 } , 6 \right)
C) (a) {−15,6}(b) (∗,−15) ∪(6,∞) ,(c) (−15,6) \left\{ - \frac { 1 } { 5 } , 6 \right\} ( b ) \left( * , - \frac { 1 } { 5 } \right) \cup ( 6 , \infty ) , ( c ) \left( - \frac { 1 } { 5 } , 6 \right)
D) (a) {−15,−1,6}\left\{ - \frac { 1 } { 5 } , - 1,6 \right\} (b) (∗,−15) ∪(6,∞) \left( * , - \frac { 1 } { 5 } \right) \cup ( 6 , \infty ) , (c) (−15,6) \left( - \frac { 1 } { 5 } , 6 \right)

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