Need help! Given y = 2x³ + bx² - cx + d. The factors (x+1), (x-2) & (2x-1) give same remainder. Wat is b & c?

Answer 1

If you put x=-1,x=2,x=1/2
you find the remainders for
the factors(x+1),(x-2),(2x-1)
respectivelly.So
For x=-1 , y=-2+b+c+d (1)
For x=2 , y=16+4b-2c+d (2)
Forx=1/2 , y=1/4+b/4-c/2+d (3)

Which are all the same
-2+b+c=16+4b-2c
-2+b+c=1/4+b/4-c/2

3b-3c=-18
3b+6c=9

c=3 and b=-3

Answer 2

Remember: the remainder of the division by
(x-a) is the value of y(a)
so using x=-1 x=2 and x=1/2
we get
r= -2+b+c+d
r=16+4b-2c+d
r=1/4+b/4-c/2 +d
Sustract the first from the second and the second from the third and you get two equation which only contain b and c
I think you can do it

Answer 3

Why not assume that the "same remainder" they give is 0, meaning that they are all roots of the cubic. To do this, multiply out you 3 binomials:

(x-1)(x-2)(2x-1)=0
(x^2-3x+2)(2x-1)=0
2x^3-x^2-6x^2+3x+4x-2=0
2x^3-7x^2+7x-2=0

By this measure, b would be -7 and c would also be -7.

Hope this helps!
-MΔTH ΔVΣNGΣR