We could do long division on this to figure stuff out. Start off with the x-1, I get the feeling it will give us immediate results. x goes into 6x^3 6x^2 times. This leaves us with a remainder of (p-6)x^2 + qx + 8. x goes into (p-6)x^2 (p-6)x times. This leaves us witha remainder of (q-p+6)x + 8. x goes into (q-p+6)x (q-p+6) times. This leaves us with a remainder of (8 - q + p - 6) = (2 - q + p). x does not go into (2 - q + p), so we are done here. The problem tells us that our remainder should equal -7. So -7 = (2 - q + p) We do long division a second time, this time with (2x-1). 2x goes into 6x^3 3x^2 times, which leaves us with a remainder of (p-3)x^2 + qx + 8. 2x goes into (p-3)x^2 ((p-3)/2)x times, which leaves us with a remainder of (q - ((p-3)/2))x + 8. 2x goes into (q - ((p-3)/2))x (q - ((p-3)/2))/2 times, which leaves us with a remainder of 8 - (q - ((p-3)/2))/2 The problem says that it is exactly divisible by (2x-1), so the remainder should actually equal zero. So 0 = 8 - (q - ((p-3)/2))/2 We now have a system of equations, and you should be able to solve for q in terms of p or p in terms of q in one or the other, substitute it into the other equation, and find the answer. Does that make sense?