Given that 2y^3+by-cy+d, where b, c and d are constants, leaves a remainder R when divided by (y+1) , (y-2) and (2y-1). Find the value of b and c. If (y+2) is a factor of the expression, find the value of d

The polynomial remainder theorem says that a polynomial p(x) leaves a remainder of p(k) when it's divided by x - k.

We're given that dividing p(y) = 2y³ + by² - cy + d leaves the same remainder R after dividing it by y + 1, y - 2, and 2y - 1. So we have

p(-1) = 2(-1)³ + b(-1)² - c(-1) + d = R

==>  R = -2 + b + c + d

p(2) = 2(2)³ + b(2)² - c(2) + d = R

==>  R = 16 + 4b - 2c + d

p(1/2) = 2(1/2)³ + b(1/2)² - c(1/2) + d = R

==>  R = 1/4 + b/4 - c/2 + d

We're also given that y + 2 is a factor, which means dividing p(y) by it leaves no remainder, and so

p(-2) = 2(-2)³ + b(-2)² - c(-2) + d = 0

==>  0 = -16 + 4b + 2c + d

Solve the system of equations in boldface. You can eliminate d from the first 3 to first solve for b and c, then solve for d :

(-2 + b + c + d) - (16 + 4b - 2c + d) = R - R

-18 - 3b + 3c = 0

b - c = -6

(-2 + b + c + d) - (1/4 + b/4 - c/2 + d) = R - R

-9/4 + 3b/4 + 3c/2 = 0

b + 2c = 3

(b - c) - (b + 2c) = -6 - 3

-3c = -9

c = 3

b - 3 = -6

b = -3

-16 + 4(-3) + 2(3) + d = 0

d = 22