Question

Express 5x³+9x²–8x–11 in the form (x–1)(x+2)Q(x) + rx + s, find the values of r and s.

Answer 1

well, if x=1, the form (x-1)(X+2)Q(x) + rx + s, or equal to -r+s
so, -5+9+8-11=-r+s
same logic if x=-2, the new form becomes (-3)(O)(Q(-2) +r(-2)+s or
5((-8))+36-16-11=-2r+s
so this leads to two equations, two unknowns, r,s. Solve.

Answer 2

I did a long division:
(5x³+9x²–8x–11) ÷ (x^2 + 2 - 2) to get

5x³+9x²–8x–11 = (x-1)(x+2)(5x+3) - x - 5

comparing this to
(x-1)(x+2)Q(x) + rx + s
so rx = -x ---> r = -1
and s = -5

I looked at Bob's solution and noticed that in the 2nd and 4th lines
-5+9+8-11=-r+s should be 5+9-8-11= r+s -----> r + s = -5
-40+36-16-11 = -2r+s should be -40+36+16-11 = -2r+s ---> -2r+s = 1
solving for r and s gave me
r = -2 and s = -3
which is not the same as mine, how do you reconcile this??

I could only conclude that my Q(x) would have to be different than Bob's Q(x)
Bob would have:
5x³+9x²-8x-11 = (x-1)(x+2)Q(x) - 2x - 3 , or
5x³+9x²-6x-8 = (x-1)(x+2)Q(x)
I then divided 5x³+9x²-6x-8 by x^2 + x -2 and got Q(x) = 5x + 4

Ahhh, so both answers are correct

Bob's: 5x³+9x²-8x-11 = (x-1)(x+2)(5x+4) - 2x - 3
Mine: 5x³+9x²-8x-11 = (x-1)(x+2)(5x+3) - x - 5

and the Q(x)'s were indeed different

and there is no unique solution!
e.g.
60 = 2*3*8 + 12
60 = 2*3*9 + 6