SOLUTION: The remainder when x^3+ax+b when divided by (x^2-1) is x+2. Find a and b.

Let me to denote / (to call) f(x) = .

The condition says that 

f(x) = g(x)*(x^2-1) + (x+2),   where g(x) is some polynomial of degree 1, but its exact expression does not matter for me now.


It is the exact meaning of this statement "The remainder of x^3+ax+b when divided by (x^2-1) is x+2".


Now, if x = 1, then

f(1)  = g(1)*(1^2-1) + (1+2) = g(1)*0 + 3 = 3.          (1)


If x = -1, then

f(-1) = g(-1)*((-1)^2-1) + (-1+2) = g(-1)*0 + 1 = 1.    (2)


In other words, using the explicit form of f(x), the equation (1) takes the form

 = 3,   or   1 + a + b = 3,   or simply  a + b = 2.           (3)


Using the explicit form of f(x), the equation (2) takes the form

 = 1,   or   -1 - a + b = 1,   or simply  -a + b = 2.     (4)


Thus you got these two equations

a + b = 2,   (5)
a - b = 2.   (6)


Add them (both sides), and you will get 2a = 4; hence, a = 2.

Then from (5),  b = 2 - a = 2 - 2 = 0.


Answer.  a = 2,  b = 0.