Question

if p(y)=4y³+2y²-y-5.then find p(0)×p(1)​

Answer 1

Answer:

above picture is the answer for your question

Answer 2

Answer:

1. p(0)= -5

2. p(1)= 0

Step-by-step explanation:

1. p(0)= 4y^3+2y^2-y-5

= 4×0^3+2×0^2-0-5

= 0+0-0-5

p(0)= -5

2. p(1)= 4y^3+2y^2-y-5

= 4×1+2×1-1-5

= 4+2-1-5

p(1)= 6-6

p(1)= 0