# Polynomial theorem: For all poly's p(x),q(x) there exists poly's k(x),r(x) such that

#### Aion

##### New member
I read in my math book today that you could prove the following theorem using long division.

For all polynomials p(x) and q(x) there exists polynomials k(x) and r(x) such that

p(x) = q(x)*k(x) + r(x)

where quotient k(x) and remainder r(x) is determined by p(x) and q(x) if the deg r(x) < deg of q(x).

My question is if anyone could show me the proof for the theorem above. Thanks.

Edit:

I've realised that my math book; which is an introductory course in mathematical analysis ( calculus 1) doesn't prove most of its theorems. Its suppose to be a book used for students in universities. But it seems to care more about maths applications than showing actual proofs. Now I don't care about its applications, I just wanna learn the underlying maths. And so I need to understand every single proof leading into calculus and beyond. Now does anyone have a recommendation for a book that proves everything that comes before and during calculus? I don't care if its hard, I just want to learn the proper way...

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#### JeffM

##### Elite Member
Let me be sure I understand what you are after.

$$\displaystyle P(x) \text { and } Q(x) \text { are polynomials } \implies$$

$$\displaystyle \exists \text { polynomials } R(x) \text { and } S(x)$$

$$\displaystyle \text {such that } P(x) = Q(x) * S(x) + R(x).$$

Is that it?

Do you really mean that the degree of R(x) > the degree of Q(x) rather than vice versa?

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#### Aion

##### New member
Yeah sorry, that was a typo lol. I meant to write deg r(x) < deg q(x).

#### Dr.Peterson

##### Elite Member
I read in my math book today that you could prove the following theorem using long division.

For all polynomials p(x) and q(x) there exists polynomials k(x) and r(x) such that

p(x) = q(x)*k(x) + r(x)

where quotient k(x) and remainder r(x) is determined by p(x) and q(x) if the deg r(x) < deg of q(x).

My question is if anyone could show me the proof for the theorem above. Thanks.
This theorem is called the "Division Algorithm" (though properly an "algorithm" is a procedure, like long division itself, by which something is done). If you search for "proof of division algorithm for polynomials" you will find it in many places, but many will be at a higher level than you need. Here is a good source for this (the first) and various other useful theorems: http://mathweb.scranton.edu/monks/courses/ProblemSolving/POLYTHEOREMS.pdf .

#### Jomo

##### Elite Member
Let me be sure I understand what you are after.

$$\displaystyle P(x) \text { and } Q(x) \text { are polynomials } \implies$$

$$\displaystyle \exists \text { polynomials } R(x) \text { and } S(x)$$

$$\displaystyle \text {such that } P(x) = Q(x) * S(x) + R(x).$$

Is that it?

Do you really mean that the degree of R(x) > the degree of Q(x) rather than vice versa?
JeffM, the OP did have the correct symbol comparing R(x) and Q(x).

#### Aion

##### New member
This theorem is called the "Division Algorithm" (though properly an "algorithm" is a procedure, like long division itself, by which something is done). If you search for "proof of division algorithm for polynomials" you will find it in many places, but many will be at a higher level than you need. Here is a good source for this (the first) and various other useful theorems: http://mathweb.scranton.edu/monks/courses/ProblemSolving/POLYTHEOREMS.pdf .
Wow, thank you so much!

#### Jomo

##### Elite Member
I read in my math book today that you could prove the following theorem using long division.

For all polynomials p(x) and q(x) there exists polynomials k(x) and r(x) such that

p(x) = q(x)*k(x) + r(x)

where quotient k(x) and remainder r(x) is determined by p(x) and q(x) if the deg r(x) < deg of q(x).

My question is if anyone could show me the proof for the theorem above. Thanks.

Edit:

I've realised that my math book; which is an introductory course in mathematical analysis ( calculus 1) doesn't prove most of its theorems. Its suppose to be a book used for students in universities. But it seems to care more about maths applications than showing actual proofs. Now I don't care about its applications, I just wanna learn the underlying maths. And so I need to understand every single proof leading into calculus and beyond. Now does anyone have a recommendation for a book that proves everything that comes before and during calculus? I don't care if its hard, I just want to learn the proper way...
Here is another video proving the division algorithm--https://www.youtube.com/watch?v=ZPtO9HMl398

I suspect that you are a math major(??). Being a math major is difficult (unless the majority of the students in the class are math majors as well) since most students (including some majors!) do not want to see proofs and as a result teachers do not too many proofs in class. The problem is that it is important for math majors to see (and do!) proofs. I commend you for noticing this problem. These days most proofs are online and you can view them. I understand that it would be best if all the calculus proofs were in one place, like your calculus textbook. Try finding an old version of Thomas/Finney calculus. They made a version with the authors labelled as Finney/Thomas--this is not the book you want.