Find the zeros of the function below algebraically, if any exist.
$$f ( t ) = \frac { 1 } { 6 } t ^ { 4 } - \frac { 27 } { 2 }$$
A) $- 3 , - 1,1$, and 3
B) $- 3$ and 3
C) $- 6$ and 6
D) $- 6 , - 1,1$, and 6
E) No zeros exist.

Question

Quizplus · Accepted Answer

Set the function equal to zero:
$$\frac{1}{6}t^4-\frac{27}{2}=0$$
$$t^4-81=0$$
$$(t^2-9)(t^2+9)=0$$
$$(t-3)(t+3)(t^2+9)=0$$
The zeroes are $t=-3$ and $t=3$, and there are no other zeroes because $t^2+9$ is always positive for real $t$. Thus, the answer is $\boxed{	extbf{(B)}}$.

Find the Zeros of the Function Below Algebraically, If Any f(t)=16t4−272f ( t ) = \frac { 1 } { 6 } t ^ { 4 } - \frac { 27 } { 2 }f(t)=61​t4−227​

Find the Zeros of the Function Below Algebraically, If Any $f ( t ) = \frac { 1 } { 6 } t ^ { 4 } - \frac { 27 } { 2 }$