# polynom remainder

• Nov 17th 2009, 10:26 AM
ferreus
polynom remainder
I get nowhere with problems of the type
p(x) gives the remainder 3 when divided by (x+3), and the remainder 4 when divided by (x-6). What is the remainder when p(x) is divided by (x+3)(x-6)?

they say it's related to the factor theorem, but i don't make the connection..
• Nov 18th 2009, 12:23 AM
Opalg
Quote:

Originally Posted by ferreus
I get nowhere with problems of the type
p(x) gives the remainder 3 when divided by (x+3), and the remainder 4 when divided by (x-6). What is the remainder when p(x) is divided by (x+3)(x-6)?

they say it's related to the factor theorem, but i don't make the connection..

I'd say that this is a case of the chinese remainder theorem for a polynomial ring.

You're told that $p(x) = q(x)(x+3)+3$ and $p(x) = r(x)(x-6)+4$, for some polynomials q(x), r(x). Multiply the first of those equations by x–6 and the second one by x+3, and subtract. You'll get $-9p(x) = s(x)(x+3)(x-6) + 3(x-6) - 4(x+3)$, for some polynomial s(x). Divide by –9 to find the remainder when p(x) is divided by (x+3)(x-6).
• Nov 18th 2009, 12:51 AM
Shanks
solve it by using the factor theorem
By the factor theoem : let p(x)=q(x)(x+3)(x-6)+ax+b,where q(x) be the quotient, a and b be undetermined coefficients.
since it leaves a remainder of 3 when divided by x+3, and a remainder of 4 when divided by x-6,we have :
p(-3)= b -3a = 3 and p(6)= b + 6a = 4
thus: a = 1/9, b =10/3.