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!!!

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.

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!

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)=xⁿ, where n ³ m (Prove that the map P(x)® 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 £ k. Now for n=k+1: P(x)=x^k+1=x×(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-1x^k-1+ terms of degree £ k-2. Then xR_k(x)=(b_k-1/a_k)g(x)+ terms of degree £ k-1. QED

Kerwin

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

Kerwin