у3 – у2 – 16у + 16 = 0±4; 01; ±41; 40; 1

Question

у3 – у2 – 16у + 16 = 0

±4; 0
1; ±4
1; 4
0; 1

Answer 1

Y²(y-1)-16(y-1)=0
(y-1)(y²-16)=0
(y-1)(y-4)(y+4)=0
y=1, y=4, y=-4