# Polynom division

1. Mar 10, 2008

### Physicsissuef

1. The problem statement, all variables and given/known data

When we divide the polynom P(x) with x-a, we receive reminder $$\alpha$$, and if we divide P(x) with x-a, we receive reminder $$\beta$$. What will be the reminder, if we divide P(x) with (x-a)(x-b)?

2. Relevant equations

3. The attempt at a solution

$$P(x)=(x-a)f(x) + \alpha$$

$$P(x)=(x-b)c(x) + \beta$$

$$P(x)= (x-b)(x-a)g(x)+ R(x)$$

We need to find R(x)

Last edited: Mar 10, 2008
2. Mar 10, 2008

### tiny-tim

Hi Physicsissuef!

Hint: write P(x) = f(x)(x-a)(x-b) + px +q.

3. Mar 10, 2008

### Physicsissuef

Yes, I already knew it. What's next? :D

4. Mar 10, 2008

### tiny-tim

Physicsissuef, you were supposed show us what you'd already done, in your original post.

Also the next thing you tried, and how it didn't work.

That's what 3. The attempt at a solution is for.​

If you don't, other PF members don't know where to start, and it wastes our time.

Start again …

5. Mar 10, 2008

### Physicsissuef

Actually I don't know why px+q
I will edited the first post. Look now. Sorry.

6. Mar 10, 2008

### tiny-tim

ok, your R(x) = px + q, for some constants p and q (because R(x) can't contain x^2, since it's a remainder after dividing by x^2 + …).

So you should be able to write two equations involving (px + q) and alpha and beta.

What are they? …

7. Mar 10, 2008

### Physicsissuef

Why after dividing x^2?

8. Mar 10, 2008

### tiny-tim

Remainders must be smaller

Actually, "after dividing by x^2 + …"

If you divide by (x^2 plus anything smaller), then the remainder has to be smaller.

If the remainder was, say, 3x^2 + 5, then you'd just subtract 3 more lots of (x^2 + …), to give you a new remainder with only x and 1.

Generally, the remainder after dividing by any polynomial beginning with x^n will be a polynomial beginning with x^(n-1) or less.

9. Mar 10, 2008

### Physicsissuef

And how do you know that it is x^2 ??

10. Mar 10, 2008

### tiny-tim

It's (x-a)(x-b), which is x^2 - (a+b)x + ab;
so when you divide any polynomial by it, the remainder will always be either linear or constant.

11. Mar 10, 2008

### Physicsissuef

And how do you know that the polynom P(x) is in this form:
$$ax^3+bx^2+cx+d$$, so you can write $$R(x)=px+q$$
??

12. Mar 10, 2008

### tiny-tim

Sorry, Physicsissuef, you've completely lost me.

Where did $$ax^3+bx^2+cx+d$$ come from?

Your own equation, $$P(x)= (x-b)(x-a)g(x)+ R(x)$$, where R(x) is a remainder, shows that R(x) can't have anything higher than x.

13. Mar 10, 2008

### Physicsissuef

You said that $$R(x)=px+q$$, right?
So when I divide some polynom with x^3 with x^2, I will receive R(x)=px+q, right?

I asked how did you know that it is x^3?

Maybe it is x^4?

14. Mar 10, 2008

### tiny-tim

Physicsissuef, divide anything with x^2 + …, you will receive R(x)=px+q.

It could be x^3 or x^4 or x^307 … it doesn't matter … the remainder will always be of the form px +q (possibly, of course, with p or q = 0).

15. Mar 10, 2008

### Physicsissuef

So what I will substitute for p and q? Sorry, if I am getting annoyed.

16. Mar 10, 2008

### tiny-tim

I've lost the plot now …

ah … we have to go back to post #6:
Hint: divide (px + q) by (x - a) … what is the remainder?

17. Mar 10, 2008

### Physicsissuef

pa+q. Why I divide them?

18. Mar 10, 2008

### tiny-tim

Because that equals alpha … can you see why?

And then, what equals beta?

19. Mar 10, 2008

### Physicsissuef

but didn't f(x)*(x-a)+px+q=P(x)

So px+q equals alpha. Because alpha is the remainder.

20. Mar 10, 2008

### tiny-tim

You've got the principle, but you've got rather confused writing it.

From post #1, P(x) = f(x)(x-a) + alpha.

And so alpha equals pa + q (not px + q), from your post #17, because, as you say, alpha is the remainder.

Right! Nearly there! alpha = pa + q.

Now what does beta equal?