# Math Help - Polynomials

1. ## Polynomials

Why is it that the most general form of the remainder in a polynomial is given by R(x) = ax +b?

2. Originally Posted by xwrathbringerx
Why is it that the most general form of the remainder in a polynomial is given by R(x) = ax +b?
Please rephrase this question, at the moment it makes no sense

RonL

3. Sori

The question was : "Explain why the most general form of R(x) i.e. the remainder in the polynomial P(x) = (x+1)(x-4) is given by R(x) = ax+b."

4. Originally Posted by xwrathbringerx
Sori

The question was : "Explain why the most general form of R(x) i.e. the remainder in the polynomial P(x) = (x+1)(x-4) is given by R(x) = ax+b."
No there is still not enough context, we have established that you are talking about second degree polynomials, but remainder for what?

RonL

5. Oops really sori it seems I've been reading the question wrong.

Explain why the most general form of R(x) is given by R(x) = ax+b.

WHEN the polynomial P(x) is divided by (x+1)(x-4) where the quotient is Q(x) and the remainder is R(x).

Sori once more

6. Originally Posted by xwrathbringerx
Oops really sori it seems I've been reading the question wrong.

Explain why the most general form of R(x) is given by R(x) = ax+b.

WHEN the polynomial P(x) is divided by (x+1)(x-4) where the quotient is Q(x) and the remainder is R(x).

Sori once more
If the remainder were a polynomial of degree greater $n$ than $1$ then it can be rewriten in the form:

$R(x)=(x+1)(x-4)P_{n-2} (x) +Q_{n-1}(x)$

Where $P_{n-2}$ and $Q_{n-1}$ are polynomials of degree $n-2$ and no more than $n-1$ respectivly, so it would not have been the remainder.