Find the zeros for the polynomial function and give the multiplicity for each zero. State whether the graph crosses
the x-axis or touches the x-axis and turns around, at each zero.
-$f ( x ) = \left( x + \frac { 1 } { 5 } \right) ^ { 2 } \left( x ^ { 2 } + 7 \right) ^ { 5 }$
A) $- \frac { 1 } { 5 }$, multiplicity 2 , touches the $x$-axis and turns around.
B) $- \frac { 1 } { 5 }$, multiplicity 2 , touches the $x$-axis and turns around;
$- 7$, multiplicity 5 , crosses the $x$-axis
C) $\frac { 1 } { 5 }$, multiplicity 2 , touches the $x$-axis and turns around;
7, multiplicity 5, crosses the $x$-axis
D) $- \frac { 1 } { 5 }$, multiplicity 2 , crosses the $x$-axis.

Question

Find the zeros for the polynomial function and give the multiplicity for each zero. State whether the graph crosses
the x-axis or touches the x-axis and turns around, at each zero.
-$f ( x ) = \left( x + \frac { 1 } { 5 } \right) ^ { 2 } \left( x ^ { 2 } + 7 \right) ^ { 5 }$
A) $- \frac { 1 } { 5 }$, multiplicity 2 , touches the $x$-axis and turns around.
B) $- \frac { 1 } { 5 }$, multiplicity 2 , touches the $x$-axis and turns around; 
$- 7$, multiplicity 5 , crosses the $x$-axis
C) $\frac { 1 } { 5 }$, multiplicity 2 , touches the $x$-axis and turns around; 
7, multiplicity 5, crosses the $x$-axis
D) $- \frac { 1 } { 5 }$, multiplicity 2 , crosses the $x$-axis.

Quizplus · Accepted Answer

The polynomial has a zero at $-\frac{1}{5}$ with multiplicity 2, which means the graph touches the x-axis and turns around at this point. The term $(x^2 + 7)^5$ does not produce real zeros because $x^2 + 7 > 0$ for all real x, so there are no additional zeros.

Find the Zeros for the Polynomial Function and Give the Multiplicity