# Thread: Remainder and Factor Theorem

1. ## Remainder and Factor Theorem

I know how to use both the remainder theorem and the factor theorem, but I'm not sure how to use them in relation to this problem:
When a polynomial is divided by $x-p$, the remainder is $p^3$.
When a polynomial is divided by $x-q$, the remainder is $q^3$.

Find the remainder when the polynomial is divided by $(x-p)(x-q)$.
Any help would be greatly appreciated, even if it's just a few hints to point me in the right direction.

2. Originally Posted by CUBE
I know how to use both the remainder theorem and the factor theorem, but I'm not sure how to use them in relation to this problem:
When a polynomial is divided by $x-p$, the remainder is $p^3$.
When a polynomial is divided by $x-q$, the remainder is $q^3$.

Find the remainder when the polynomial is divided by $(x-p)(x-q)$.
Any help would be greatly appreciated, even if it's just a few hints to point me in the right direction.
Hello, maybe you are looking for this?: http://www.mathhelpforum.com/math-he...lynomials.html

3. I don't understand the method used in that thread. Could somebody else explain it better?

4. Originally Posted by CUBE
I know how to use both the remainder theorem and the factor theorem, but I'm not sure how to use them in relation to this problem:
When a polynomial is divided by $x-p$, the remainder is $p^3$.
When a polynomial is divided by $x-q$, the remainder is $q^3$.

Find the remainder when the polynomial is divided by $(x-p)(x-q)$.
Any help would be greatly appreciated, even if it's just a few hints to point me in the right direction.
I can't think of a less complex method to use. Perhaps you just need a little more explanation red_dog's method.

You are given a polynomial f(x). You know that when you divide f(x) by x - p the remainder is $p^3$. So you know that
$f(x) = (x - p)g_1(x) + p^3$
by the division theorem, where $g_1(x)$ is some polynomial. By the remainder theorem (or just plain substitution) we also see that
$f(p) = (p - p)g_1(p) + p^3 = p^3$

Now, f(x) divided by x - q leaves a remainder of $q^3$. So by exactly the same line of reasoning we have that
$f(x) = (x - q)g_2(x) + q^3 \implies f(q) = q^3$
where $g_2(x)$ is another polynomial.

Now we wish to divide f(x) by (x - p)(x - q). Now, we are dividing by a quadratic polynomial. When we do this in general we are not left with a constant remainder, we are left with a remainder polynomial of one degree less than the divisor: we are left with a first degree polynomial as the remainder. So f(x) must be of the form:
$f(x) = (x - p)(x - q)g_3(x) + (ax + b)$
where $g_3(x)$ is another polynomial.

Possibly you would be more comfortable with this format?
$\frac{f(x)}{(x - p)(x - q)} = g_3(x) + \frac{ax + b}{(x - p)(x - q)}$
This is the same statement.

Plugging x = p and x = q into this equation gives us:
$f(p) = (p - p)(p - q)g_3(p) + (ap + b) = ap + b$
and
$f(q) = (q - p)(q - q)g_3(q) + (aq + b) = aq + b$

But we already know that $f(p) = p^3$ and $f(q) = q^3$ so we have the system of equations:
$ap + b = p^3$
$aq + b = q^3$

Now solve for a and b:
$a = p^2 + pq + q^2$
$b = -pq(p +q)$

(Note that if $p = q$ we have no solution for a and b, so the problem really needed so specify that $p \neq q$.)

So when f(x) is divided by (x - p)(x - q) we get a remainder of $(p^2 + pq + q^2)x - pq(p + q)$.

If you want a concrete example of all this, consider $f(x) = x^3 + 2x^2 - 2x - 4$ and p = -1 and q = 2.

-Dan