Find all the zeros of the function and write the polynomial as a product of linear fac- tors.
$y ^ { 4 } - 2401$
A) $\pm 7 ( y - 7 ) ( y + 7 )$
B) $\pm 7 , \pm 7 i ; ( y - 7 ) ( y + 7 ) ( y - 7 i ) ( y + 7 i )$
C) $7,7 i ; ( y + 7 ) ( y + 7 ) ( y - 7 i ) ( y - 7 i )$
D) $\pm 7 i ; ( y - 7 i ) ( y + 7 i )$
E) $\pm 7 , \pm 7 i ; ( y - 7 ) ( y - 7 ) ( y + 7 i ) ( y + 7 i )$

Question

Find all the zeros of the function and write the polynomial as a product of linear fac- tors. 
$y ^ { 4 } - 2401$ 
A) $\pm 7 ( y - 7 ) ( y + 7 )$
B) $\pm 7 , \pm 7 i ; ( y - 7 ) ( y + 7 ) ( y - 7 i ) ( y + 7 i )$
C) $7,7 i ; ( y + 7 ) ( y + 7 ) ( y - 7 i ) ( y - 7 i )$
D) $\pm 7 i ; ( y - 7 i ) ( y + 7 i )$
E) $\pm 7 , \pm 7 i ; ( y - 7 ) ( y - 7 ) ( y + 7 i ) ( y + 7 i )$

Quizplus · Accepted Answer

To find the zeros of the function, we can set it equal to zero and solve for y:

$$y^4-2401=0$$
$$(y^2)^2-(49)^2=0$$
$$(y^2-49)(y^2+49)=0$$

So the zeros are $y= \pm 7$ and $y= \pm 7i$.

To write the polynomial as a product of linear factors, we can use these zeros. The factors corresponding to the real zeros are simply $(y-7)$ and $(y+7)$. The factors corresponding to the complex zeros are $(y-7i)$ and $(y+7i)$.

Therefore, the polynomial can be expressed as:

$$(y^2-49)(y^2+49)=(y-7)(y+7)(y-7i)(y+7i)$$

So the best choice is B, which correctly identifies the zeros and the corresponding linear factors.

Find All the Zeros of the Function and Write the Polynomial