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There are two quadratic polynomials (dividends). These two polynomials are divided by two different linear polynomials like $x+1$ (divisors). The remainders are known, but the quotients are unknown.

For example: $$\dfrac{ax^2 + bx + 1}{x+1} \text{ gives remainder } 3$$ and $$\dfrac{bx^2 + ax + 2}{x-2} \text{ gives remainder } 2$$

I have been thinking about the Remainder Theorem. However, since there are two unknowns in the quadratic polynomials plus the unknown quotients, it seems we have fewer simultaneous equations than unknowns to solve for.

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  • $\begingroup$ @hardmath I apologise. Corrected. $\endgroup$ – Johnson Dec 8 '13 at 12:58
  • $\begingroup$ I think you are on track with the Remainder Theorem. Since the remainders are constants, you really do know the quotient part of the (polynomial) divisions (because the leading coefficients have to get eliminated). $\endgroup$ – hardmath Dec 8 '13 at 13:01
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The Remainder Theorem, usually presented in connection with Synthetic Division, tells us that dividing a polynomial $p(x)$ by a monic first-degree polynomial $x-r$ gives us a (constant) remainder $p(r)$, the polynomial $p(x)$ evaluated at $x=r$.

So the first relation tells us about $ax^2 + bx + 1$ being evaluated at $x=-1$ (we should get $3$), and the second relation says $bx^2 + ax + 2$ evaluated at $x=2$ should get $2$.

Now you have two equations in two unknowns, $a$ and $b$, so there's a chance this gives a unique solution.

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  • $\begingroup$ Thank you very much. Your answer is clear and you have taken my eye-mask off. $\endgroup$ – Johnson Dec 8 '13 at 13:24

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