Polynomial division and remainders

By Anonymous on Monday, April 30, 2001 - 12:51 am :

Please help with this problem:

A polynomial p(x) has remainder of 7 when divided by x-2 and remainder of 1 when divided by x+3. Find the remainder when p(x) is divided by (x-2)(x+3).
My book says the remainder is a linear one. Is it necessarily so?

Thanks!!!

By Michael Doré (Michael) on Monday, April 30, 2001 - 06:05 pm :

Yes, that's right. Whenever you divide a polynomial of degree n by a polynomial of degree m, the remainder is always a polynomial of degree m-1 (would you like a proof?)

So p(x) = (x - 2)(x - 3)q(x) + r(x)

where r(x) is linear.

But by the remainder theorem, p(2) = 7 and p(-3) = 1 so by substituting into (*):

7 = r(2)
1 = r(-3)

But r(x) = ax + b, so -3a + b = 1 and 2a + b = 7
So a = 6/5 and b = 23/5 and the remainder is 6/5x + 23/5.

By Yiingleong Chin (P3505) on Monday, May 21, 2001 - 10:53 am :

Hi!

I am interested in the proof that says whenever you divide a polynomial of degree n by a polynomial of degree m, the remainder is always a polynomial of degree m-1. Can you, or anyone show me how to do that?
Thanks a lot!

By Kerwin Hui (Kwkh2) on Monday, May 21, 2001 - 02:15 pm :

Yiingleong,

We define the remainder $R (x)$ when a polynomial $P (x)$ is divided by another polynomial $g (x)$ by:

$P (x) = Q (x) g (x) + R (x)$

where $Q (x)$ is a polynomial and $R (x)$ is a polynomial of degree less than $\deg (g)$ .

We can always do this. It suffices to show that we can do so for $P (x) = x^{n}$ , where $n \geq m$ (Prove that the map $P (x) \to R (x)$ is linear). We shall do so by induction.

Basic Step: If $P (x) = x^{m}$ , and $g (x) = a_{m} x^{m} +$ lower order terms. Then, set $Q (x) = 1$ and we see that $R (x)$ contains only terms of degree less than $m$ .

Induction Step: Suppose the hypothesis holds true for $n \leq k$ . Now for $n = k + 1$ : $P (x) = x^{k + 1} = x \times (x^{k})$ . Now,

$x^{k} = Q_{k} (x) g (x) + R_{k} (x)$

so $x^{k + 1} = {xQ}_{k} g (x) + {xR}_{k} (x)$ .

Soif $R_{k}$ is of degree $< m - 1$ , ${xR}_{k} (x)$ is the remainder term $R_{k + 1}$ .

On the other hand, if $R_{k} (x)$ is of degree $k - 1$ , say $R_{k} (x) = b_{k - 1} x^{k - 1} +$ terms of degree $\leq k - 2$ . Then ${xR}_{k} (x) = (b_{k - 1} / a_{k}) g (x) +$ terms of degree $\leq k - 1$ . QED

Kerwin

By Kerwin Hui (Kwkh2) on Monday, May 21, 2001 - 02:22 pm :

Note also that the remainder is of degree less than m, but not necessarily of degree m-1.

Kerwin