# Dividing Polynomials ~ Root/Factor Theroem

1. Oct 18, 2007

1. The problem statement, all variables and given/known data

A cubic polynomial gives remainders $$(5x+4)$$ and $$(12x-1)$$ when divided by $$x^{2} - x + 2$$ and $$x^{2} + x - 1$$ respectively. Find the polynomial

2. Relevant equations
:S Well, im using the root theorem, the factor theorem, and possibly just basics on long division..

We know that:

$$\frac{P(x)}{D(x)}=Q(x) + \frac {R(x)}{D(x)}$$

3. The attempt at a solution

$$P(x)=ax^{3} + bx^{2} + cx + d$$ which is our standard for a cubic polynomial.

**Please note I will use Q(x1) and Q(x2) for the tw different quotients of the two divisions.
Then we know that

$$\frac {P(x)}{x^{2} - x + 2} = Q (x_{1}) + \frac{5x+4}{x^{2}-x+2}$$

Therefore, our P (x) is the following for the first division.

$$P(x)=Q(x_{1})(x^{2} - x + 2) + 5x + 4$$

If we do the same for the next division, we obtain the following (using the same procedure)

$$P(x)=Q(x_{2})(x^{2} + x + 1) + 12x - 1$$

I have no idea, I can possibly make them equal each other, and sort of solve:
$$Q(x_{2})(x^{2} + x + 1) + 12x - 1=Q(x_{1})(x^{2} - x + 2) + 5x + 4$$

$$Q(x_{2})(x^{2} + x + 1) =Q(x_{1})(x^{2} - x + 2) - 7x + 5$$

$$Q(x_{2})(x^{2} + x + 1) =Q(x_{1})(x^{2} - x + 2) - 7x + 5$$

...any advice to lead me into the right path, I feel im going the wrong way.

Last edited: Oct 18, 2007
2. Oct 19, 2007

### qspeechc

Q(x) has to be linear. So put Q(x) = px+q, say.

Let the cubic be ax^3 + bx^2 +cx +d
ax^3 + bx^2 +cx +d = (px+q)(x^2 - x + 2) + 5x +4

ax^3 + bx^2 + (c-5)x + (d-4) = (px + q)(x^2 - x + 2)

And similarly for the other one. Then it is solvable by equation like powers yadda yadda yadda.