Question

find all the zeros of the function. (enter your answers as a comma - separated list.)
$f(y)=81y^{4}-16$
$y = $
write the polynomial as a product of linear factors. use a graphing utility to verify your results graphically. (if possible, use the
$f(y) = $
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Explanation:

Step1: Recognize difference of squares

$f(y) = (9y^2)^2 - 4^2$

Step2: Factor first difference of squares

$f(y) = (9y^2 - 4)(9y^2 + 4)$

Step3: Factor remaining difference of squares

$9y^2 - 4 = (3y)^2 - 2^2 = (3y-2)(3y+2)$

Step4: Factor sum of squares (complex)

$9y^2 + 4 = (3y)^2 - (2i)^2 = (3y-2i)(3y+2i)$

Step5: Set $f(y)=0$ to find zeros

$3y-2=0 \implies y=\frac{2}{3}$; $3y+2=0 \implies y=-\frac{2}{3}$; $3y-2i=0 \implies y=\frac{2}{3}i$; $3y+2i=0 \implies y=-\frac{2}{3}i$

Answer:

$y = -\frac{2}{3}, \frac{2}{3}, -\frac{2}{3}i, \frac{2}{3}i$
$f(y) = (3y-2)(3y+2)(3y-2i)(3y+2i)$