Question

finding zeros of polynomial
8y2 - 3y​

Answer 1

Answer:

Step-by-step explanation:

8y^2-3y

y(8y-3)=0

y=0 ; 8y-3=0

y=3/8

alpha=0 & beta=3/8

alpha+beta=0+3/8= -(-3/8)= -b/a

alpha*beta=(0)(3/8)=0=c/a