$$f ( x ) = 5 \left( x ^ { 2 } + 6 \right) \left( x ^ { 2 } + 4 \right) ^ { 2 }$$
A) $\sqrt { 6 }$, multiplicity 1 , crosses $x$-axis; $- \sqrt { 6 }$, multiplicity 1 , crosses $x$-axis; 2 , multiplicity 2 , touches $x$-axis; $- 2$, multiplicity 2 , touches $x$-axis
B) -6, multiplicity 1 , touches $x$-axis; $- 4$, multiplicity 2, crosses $x$-axis
C) -6, multiplicity 1 , crosses $x$-axis; $- 4$, multiplicity 2 , touches $x$-axis
D) No real zeros

Question

Quizplus · Accepted Answer

The function $f(x)$ is a product of two quadratic expressions, each of which has no real roots. Therefore, $f(x)$ has no real zeros.

f(x)=5(x2+6)(x2+4)2f ( x ) = 5 \left( x ^ { 2 } + 6 \right) \left( x ^ { 2 } + 4 \right) ^ { 2 }f(x)=5(x2+6)(x2+4)2

$f ( x ) = 5 \left( x ^ { 2 } + 6 \right) \left( x ^ { 2 } + 4 \right) ^ { 2 }$