Let $f(x)$ be a cubic polynomial. If $f(x)$ is devided by $x+2$, the remainder is $-10$. If f(x) is divide by (x-1), the remainder is 20.

(a) If $f(x)$ is divided by $x^2+x-2$, find the remainder.

(b) If $f(x)$ is divided by $2-x-x^2$, will the remainder the same as the answer in (a)?

I do not know how to solve the question without the polynomial formula, so there I am asking for help.

  • 4
    $\begingroup$ Are you sure this is the question? It seems to me we don't have enough information. We see that $x^2+x-2 = (x-1)(x+2)$ but we only know the remainder if we divide by $x+2$ and nothing about $x-1$. $\endgroup$
    – Jolien
    Apr 10 '15 at 8:17
  • $\begingroup$ you 're right so. $\endgroup$
    – dick hui
    Apr 10 '15 at 8:53

For (a), you have to solve the system of congruences $$\tag{sys} \begin{cases} f \equiv -10 \pmod{x + 2}\\ f \equiv 20 \pmod{x - 1}. \end{cases} $$ This is because a solution of this will be of the form $$ f = f_{0} \pmod{(x + 2) (x -1)}, $$ where $f_{0}$ is a particular solution of (sys), and $(x+2)(x-1)$ is the lcm of $x+2$ and $x-1$, as their gcd is $1$.

There is a standard way for finding $f_{0}$. Find a linear combination of $x + 2$ and $x - 1$ that yields their gcd, which is $1$, or a nonzero scalar multiple thereof: we choose $3$

So you may take $$ (x + 2) - (x - 1) = 3.$$

Multiply by $10$ to get $$ 10 (x + 2) - 10 (x - 1) = 30 = 20 - (-10), $$ and thus $$\tag{sol} 10 (x + 2) - 10 = 10 (x -1) + 20 = 10 x + 10, $$ and this will be the requested remainder $f_{0}$. Just check using (sol) that any polynomial of the form $$ (x^{2} + x - 2) g + 10 x + 10 $$ for any $g$, will have the correct remainders modulo $x+2$ and $x-1$.

The fact that $f$ is cubic appears to be irrelevant. Also, the answer to (b) is yes.


Note that polynomial division of $p(x)$ by $d(x)$ gives $$p(x)=d(x)q(x)+r(x)$$ where $q(x)$ is a polynomial and $r(x)$ has degree less than the degree of $d(x)$. This requires that the polynomials be defined over a field - rationals, reals or complex numbers will do. In such cases $r(x)$ is unique.

We can use this with $d(x)=x+2$ to give $p(x)=(x+2)q(x)+A$, (because $r(x)$ has to be constant) and $p(-2)=-10$. Similarly $p(1)=20$.

Now to the problem itself. Here we do not know $q(x)$ or $r(x)$, but we can write $$p(x)=(x^2+x-2)q(x)+ax+b$$

(the degree of the remainder is at most $1$). If we set, respectively $x=-2, x=1$ we find $$p(-2)=-10=-2a+b$$ and $$p(1)=20=a+b$$

And from this we see $a=b=10$.

It is also easy to check what happens when you change the sign of $r$.


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