Chapter 3 - 3.4 - Zeroes of Polynomial Functions - 3.4 Exercises - Page 284: 66

$f(y)=(y+4)(y-4)(y+4i)(y-4i)$ The zeros: $y=-4$ $y=4$ $y=-4i$ $y=4i$

$f(y)=y^4-256$ $f(y)=(y^2)^2-16^2$ $f(y)=(y^2-16)(y^2+16)$ $f(y)=(y^2-4^2)[y^2-(4i)^2]$ $f(y)=(y+4)(y-4)(y+4i)(y-4i)$ The zeros: $y+4=0$ $y=-4$ $y-4=0$ $y=4$ $y+4i=0$ $y=-4i$ $y-4i=0$ $y=4i$