Question

Factor and remainder theorem:
Divide f(x)=2x^3+5x^2-7x-10 by x+2
Remainder = f(-2) =8
Quotient=
(2x^3+5x^2-7x-10)/(x+2)=ax^2+bx+c
2x^3+5x^2-7x-10=(x+2)(ax^2+bx+c)
2x^3+5x^2-7x-10=ax^3+2ax^2+bx^2+2bx+cx+2c
2x^3+5x^2-7x-10=ax^3+(2a+b)x^2+(2b+c)x+2c
Equate coefficients:
a=2
2a+b=5, 4+b=5, b=1
2b+c=-7, 2+c=-7, c=-9 <-
2c=-10, c=-5 <-

Why is there two different values for c when equating the coefficients?