Remainder theorem only works with quadratics divided by linear?

[zeion]

Homework Statement

The remainder theorem can't really be applied when dividing by something other than a linear equation since you wouldn't know what a is, right?

Homework Equations

The Attempt at a Solution

 

[icystrike]

Why not? your r(x) may not neccessarily be just a real number it can be a function in term of x.

 

[Mark44]

f(x) can be a polynomial of any degree, not just degree 2 (quadratic). And the divisor can be a polynomial of any degree up to the degree of f(x).

Note that you are not dividing by a linear (or other) equation; you are dividing by a polynomial expression.

 

[zeion]

So how do I know what the a is in the divisor if its not linear?

 

[symbolipoint]

So how do I know what the a is in the divisor if its not linear?

We are taught to use synthestic division for using the Remainder Theorem, but we do not need to rely on just that. Were we taught synthetic division for dividing by a quadratic expression? The Remainder theorem is intended for examining possible binomial linear factors of a polynomial function; sometimes you will find that factoring might give or show quadratic factors. Maybe my impression is wrong about that; the impression comes from not recalling using the Remainder Theorem when looking at possible quadratic factors for a function, and not recalling ever having used synthetic division to divide by a quadratic expression. Try creating an example if you want.

 

[symbolipoint]

Post #2 should be enough. Having an equation of polynomials does not place any specification of using synthetic division for examining binomial factors. If any general f(x) is of degree 4 or higher, there may be some f(x) which could be factorable into two or more quadratic factors.

 

[statdad]

The original question was about the remainder theorem which refers to the divisor having the form $$x - a$$ for some constant $$a$$. It states that the remainder when $$f(x)$$ is divided by $$x - 1$$ is equal to $$f(a)$$. In that sense, the original poster is correct: the remainder theorem applies only in this case.

 

[HallsofIvy]

Good- with the provision that f must be a polynomial, of course.

 

[statdad]

Good- with the provision that f must be a polynomial, of course.

Correct - I should have stated this as well. I also had typo: my comment should refer to division by $$x - a$$ rather than division by $$x - a$$.

 

[zeion]

If f(x) was of degree n and it is divided by (x-a) then f(a) would give me r(a) where r(x) is a polynomial of degree n-1, right?

Is there a way to find what r(x) is?

 

[statdad]

If f(x) was of degree n and it is divided by (x-a) then f(a) would give me r(a) where r(x) is a polynomial of degree n-1, right?

Is there a way to find what r(x) is?

If you divide

$$f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0$$

by [itex]x - c [/tex]<br /> <br /> you can write things as<br /> <br /> $$f(x) = q(x) (x-c) + r$$<br /> <br /> where $r = f(c)$ is a constant. It is the quotient $q(x)$ that has degree $n - 1$, and you can find it (its coefficients, actually) with the process of synthetic division or long division.[/itex]

 

[zeion]

How come the remainder is written as r(x) if it is always a constant and does not involve x?

 

[symbolipoint]

The degree of f(x) is greater than or equal to the degree of the q(x)g(x), assuming that g(x) is used as a binomial divisor. The Remainder Theorem relies on polynomial division, not Real Number division. The remainder which may occur is a function, not just a number.

 

[HallsofIvy]

zeion, there has been confusion throughout this thread as to exactly what you meant by the "remainder theorem".

The general "division algorithm" for polynomials says that if P(x) is a polynomial and D(x) is a polynomial of lower degree than P, then P(x)/Q(x)= D(x)+ R(x)/Q(x) Where R(x) has degree less than the dergree of Q, or, as icystrike put it in his first response, P(x)= D(x)Q(x)+ R(x).

But the "remainder theorem" as statdad said, says that if P(x) is a polynomial and a is a number, then P(x)= Q(x)(x- a)+ P(a) where P(a), the value of the polynomial P(x) at x= a, is a constant, NOT a function of x.