Question

p(y) = 2y³ + y² + 6y when y =1 and y = -1 ​

Answer 1

Answer:

2y^3+y^2-2y-1

let p(y) = 2y^3+y^22y^3+y^2-2y-1

By trial,we find that

p(1)=2(1)^3+(1)^2-2(1)-1

= 2 + 1 - 2 -1=0

By factor theorem,(y -1)is a factor of p(y)

Now

2y^3+y^22y^3+y^2-2y-1

= 2x^2(y-1) +3y(y-1)+1(y-1)

= (y-1) (2y^2+3y+1)

= (y-1) (2y^2+2y+1)

= (y-1) (2y(y+1) +1 (y+1))

= (y-1) (y+1) (2y+1).

Answer 2

Step-by-step explanation:

p(y) = 2y³ + y² + 6y

When y = 1

p(1) = 2(1)³ + (1)² + 6×1

= 2×1 +1 + 6

= 2 + 1 + 6

= 9

When y = -1

p(-1) = 2(-1)³ + (-1)² + 6×-1

= 2×-1 + 1 -6

= -2 + 1 - 6

= -8 + 1

= -7