Polynom Division: Finding the Reminder with (x-a)(x-b)

[Physicsissuef]

Homework Statement

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)?

Homework Equations

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)

 

[tiny-tim]

Homework Statement

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)?

The Attempt at a Solution

Hi Physicsissuef!

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

 

[Physicsissuef]

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

 

[tiny-tim]

Yes, I already knew it.

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

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 …

 

[Physicsissuef]

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

 

[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? …

 

[Physicsissuef]

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? …

Why after dividing x^2?

 

[tiny-tim]

Remainders must be smaller

Why after dividing x^2?

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.

 

[Physicsissuef]

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

 

[tiny-tim]

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

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.

 

[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
??

 

[tiny-tim]

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
??

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.

 

[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?

 

[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).

 

[Physicsissuef]

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

 

[tiny-tim]

I've lost the plot now …

ah … we have to go back to post #6:

ok, your R(x) = px + q, for some constants p and q. {snip}

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

What are they? …

Hint: divide (px + q) by (x - a) … what is the remainder?

 

[Physicsissuef]

pa+q. Why I divide them?

 

[tiny-tim]

pa+q. Why I divide them?

Because that equals alpha … can you see why?

And then, what equals beta?

 

[Physicsissuef]

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

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

 

[tiny-tim]

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

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

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?

 

[Physicsissuef]

pb+q

 

[tiny-tim]

oops!

oops! my last post was wrong!

should have been:

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

Suppose the remainder on dividing f(x) by (x-b) is B.

Then the remainder on diving P(x) by (x-a)(x-b) will be alpha plus B(x-a).

In other words: px + q = alpha plus B(x-a), for some constant B.

Similarly, px + q = beta plus A(x-b), for some constant A.

So alpha plus B(x-a) = beta plus A(x-b)

So … ​

Sorry!

 

[Physicsissuef]

Man, why dividing f(x) with (x-b)?

 

[tiny-tim]

Because the remainder on diving P(x) by (x-a)(x-b) is px + q.

So suppose f(x) = g(x)(x-b) + B, for some polynomial g(x), and some constant (remainder) B.

So P(x) = f(x)(x-a) + alpha
= (g(x)(x-b) + B)(x-a) + alpha
= g(x)(x-b)(x-a) + B(x-a) + alpha,
and so the remainder on dividing P(x) by (x-b)(x-a) is B(x-a) + alpha.

So px + q = B(x-a) plus alpha, for some constant B.

 

[Physicsissuef]

and why in my textbook, the result is
R(x)=mx+n= \frac{\alpha - \beta}{a-b}x + \frac{b\alpha - \alpha\beta}{a-b}

??

 

[tiny-tim]

That's correct! (except it's a\beta, not \alpha\beta, at the end)

You tell me why! ​

 

[Physicsissuef]

That's correct! (except it's a\beta, not \alpha\beta, at the end)

You tell me why! ​

I don't know why... Please helllppp

 

[tiny-tim]

ok, you already have

px + q = B(x-a)\,+\,\alpha\,.​

And similarly you can get

px + q = A(x-b)\,+\,\beta\,.​

So

B(x-a)\,+\,\alpha\,=\,A(x-b)\,+\,\beta\,.​

So what do B and A equal?

(Remember, you know the result is
R(x)=mx+n= \frac{\alpha - \beta}{a-b}x + \frac{b\alpha - \alpha\beta}{a-b};
of course the book is using m and n instead of my p and q.)

 

[Physicsissuef]

B= \frac{A(x-b)}{x-a}+ \frac{\beta - \alpha}{x-a}

A= \frac{B(x-a)}{x-b}+ \frac{\alpha - \beta}{x-b}

What is next?

 

[tiny-tim]

Physicsissuef, you're really bad at this!

Don't do any dividing!

Just look at:

B(x-a)\,+\,\alpha\,=\,A(x-b)\,+\,\beta\,.​

You have a polynomial on the left, and another one on the right.

You want these polynomials to be equal (for all values of x).

So …

 

[Physicsissuef]

B=A
-B*a+ alpha =-A*b+ beta

Like this?

 

[tiny-tim]

Yes!

So the remainder = px + q = mx + n = … ?

 

[Physicsissuef]

=-A*a+alpha=-A*b+beta
what's next? :) :)

 

[tiny-tim]

=-A*a+alpha=-A*b+beta

Right!

-A*a+alpha=-A*b+beta​

a b alpha and beta were given in the original question.

Your only unknown now is A (now we've got rid of that irritating B )

So A = … ?

 

[Physicsissuef]

A= \frac{\alpha - \beta}{a-b}

What is next? LOL

 

[tiny-tim]

You're virtually there!

You're just round the corner! (has anyone ever told you that before? …)

px + q = A(x-b)\,+\,\beta

A= \frac{\alpha - \beta}{a-b}​

So the remainder = px + q = mx + n = … ?

 

[Physicsissuef]

\frac{x(\alpha - \beta)}{a-b}+\frac{\beta a-\alpha b}{a-b}
I substitute for B(x-a) + alpha.

 

[tiny-tim]

Hurrah!

Case closed?

 

[Physicsissuef]

Hurrah!

Case closed?

I think so. Thanks buddy

 

[Physicsissuef]

Just, want to ask you, we have:
\frac{\beta a-\alpha b}{a-b}
and in my textbook result:
\frac{\alpha b - \beta a}{a-b}

Is their fault?

 

[tiny-tim]

Hi Physicsissuef!

I've gone over it again, and I can't see any mistakes.

Let's test it with a = 2, b = 1, P(x) = x^2\,-\,2x\,+\,3.

Then alpha = 3, beta = 2.

And (x-a)(x-b) = (x-2)(x-1) = x^2\,-\,3x\,+\,2, so R(x) = x + 1.

\frac{x(\alpha - \beta)}{a-b}+\frac{\beta a-\alpha b}{a-b}

= (3-2)x/(2-1) + (2.2 - 3.1)/(2-1)

= x + 1.

So our formula is right, and the textbook is wrong!

Hurrah!