Question

2x³+bx²-cx+d leaves the same remainder, when divided by x+1 (or)x-2(or)2x-1. Find b and c.

Answer 1

F(x) = 2x^3+bx^2-cx+d
Since x+1= 0, x= -1.
f(-1) = 2(-1)^3+ b(-1)^2-c(-1)+d
= -2+b+c+d

Since x-2=0, x=2
f(2)= 2(2)^3+b(2)^2-c(2)+d
= 16+4b-2c+d

Since 2x-1 =0, x= 0.5
f(0.5)= 2(0.5)^3+b(0.5)^2-c(0.5)+d
= 0.25 + 0.25b - 0.5c+d

As it has the same remainder, i can conclude that:
-2+b+c+d=16+4b-2c+d=0.25+ 0.25b - 0.5c+d
-2+b+c=16+4b-2c=0.25+ 0.25b - 0.5c
break them up into 2 equality.
-2+b+c=16+4b-2c -------------------(1)
and
16+4b-2c=0.25+ 0.25b - 0.5c-----------------(2)


From (1)
-2+b+c=16+4b-2c
b+c= 18+4b-2c
0=18+3b-3c
3c=18+3b
c= 6+b-----------------(3)

From (2)
16+4b-2c=0.25+ 0.25b - 0.5c
64+16b-8c= 1+b-2c
63+15b= 6c-------------(4)
Substitute (3) into (4)
63+15b= 6( 6+b)
63+ 15b= 36+6b
9b= -27
b= -3
Substitute b= -3 into (3)
c= 6-3
c=3

Therefore, 2x^3+bx^2-cx+d => 2x^3 -3x^2-3x+ d
Since x+2=0, x= -2
f(-2) = 2x^3 -3x^2-3x+ d
= 2(-2)^3 -3(-2)^2-3(-2)+ d
= -16 -12 +6+d

As f(x) is divisible by x+2, it implies that
-16 -12 +6+d = 0
d=22

Therefore b= -3, c=3, d= 22