How do I compute for a cubic polynomial which leaves a remainder of (x-2) when divided by (x-1) ²and a remainder of (2x+2) when divided by (x²-2x)?

Answer 1

It's a bit tedious but he's how:

1) write the generic 3rd order polynomial as ax³ + bx² + cx + d

2) multiply out (x - 1)² to be x² - 2x +1 and divide it into the above using long division. You'll get a remainder that says: (3a + 2b + c)x + (-2a - b + d) but since the remainder is given in the problem as (x - 2) you set the x coefficient "(3a + 2b + c)" equal to the x coefficient of the given remainder (1) and you have the first linear equation. 3a + 2b + c = 1. Do the same with the constant term, -2a - b + d = -2

3) Divide x² - 2x into ax³ + bx² + cx + d to get the other two linear equations, 4a + 2b + c = 2 and d = 2.

4) stick these into your calculator to solve them simultaneously (or solve by hand) to get a = 1, b = 2, c = -6, and d = 2

5) Plug these in to get the cubic polynomial: x³ + 2x² - 6x + 2