Find polynoms, with as least as power possible

[Theofilius]

Homework Statement

Find polynoms, with as least as power possible, so they will be competent for the equation.

$$(x^4+2x^3+x+1)A(x) + (x^4+x^3-2x^2+2x-1)B(x)=x^3-2x$$

Homework Equations

The Attempt at a Solution

 

[Theofilius]

Please help.

 

[tiny-tim]

Hi Theofilius!

Just use trial and error …

A and B scalars obviously doesn't work.

So try A and B first-order, of the form A(x) = ax + b, B(x) = cx + d; if that doesn't work, try second-order, if that doesn't work …

Good luck!

 

[Theofilius]

Is there another way, much sure than this one ? :)

 

[HallsofIvy]

There are two ways to do it- you can wave a magic wand or you can do the work! Why would you consider what Tiny Tim suggested "not sure"?

A little thought might simplify the work: since there are no x⁴, x² or constant terms on the right hand side, those will have to cancel out on the left.

 

[Theofilius]

I mean, if it is $$x^100 + x^99$$, aren't any fixed way? If not its ok.

 

[tiny-tim]

I mean, if it is $$x^100 + x^99$$, aren't any fixed way? If not its ok.

(erm … you have to use {} round the 100 or the 99, or it comes out like … well, like that!)

Sorry, Theofilius, not understanding you … not even with the babel fish.

 

[Theofilius]

Sorry, I meant if there any 100% way sure, is there any fixed way, so I can solve the equation withouth trying?

 

[tiny-tim]

Sorry, I meant if there any 100% way sure, is there any fixed way, so I can solve the equation withouth trying?

:rofl: Theofilius, nice try! :rofl:​

… but there wouldn't be any point in your teachers setting these exercises if you could do them without trying, would there?

 

[Theofilius]

Sorry, if I am misunderstud. I want to know is there any fixed principle for solving this equation? If not, its ok. In which form will be A(x). $$A(x)=ax^2+bx+c$$. What about B(x)?

 

[tiny-tim]

Sorry, if I am misunderstud. I want to know is there any fixed principle for solving this equation? If not, its ok. In which form will be A(x). $$A(x)=ax^2+bx+c$$. What about B(x)?

So far as I know, the "fixed principle" is the trial-and-error method I suggested.

(It may also be possible to do it by dividing by $$x^3\,-\,2x$$ and then dealing with the remainders, but
(a) I think that might take even longer for a relatively short problem like this, and
(b) you're not good at dividing by polynomials, are you?)

Try A(x) and B(x) both of the form ax + b first.

If that doesn't work, try A(x) and B(x) both of the form $$ax^2+bx+c$$.

If that doesn't work, …

 

[Theofilius]

I tried with dividing, but it didn't worked. I will try with the other method.

 

[Theofilius]

Ok, I tried to solve the system of equation, and it didn't work.
$$A(x)=ax^2+bx+c$$ $$B(x)=ex^2+fx+g$$
Also I tried with
$$A(x)=ax^2+bx+c$$ $$B(x)=ax^2+bx+c$$
and nothing.

 

[Theofilius]

There are 1000 of combinations, isn't any simpler way of doing this job?

 

[tiny-tim]

You're almost there …

Theofilius, you're giving up too easily …

You're nearly half-way there, because your own equations immdiately eliminate some of the unknowns.

You've written a + e = 0 and c - g = 0.

So e = -a, and g = c.

Now rewrite the other four equations, without e and g (for example the second line is b + a + f = 0) …

 

[Physicsissuef]

it has no solution

 

[Physicsissuef]

Really if some polynom is given, we should try (ax+b) for the 1st one, and then (cx+d) for the second one. If not, then (ax^2+bx+f), and for the 2nd one (cx+d) and vice versa, which is too complicated, and wasting time, there must be some other solution, less complicated than this one.

 

[Theofilius]

I hope so that there is another way, because if it is given on some paper test, it will take me 30min. to solve it.

 

[Physicsissuef]

By dividing the whole polynom with $x^3-2x$, we receive:

$$((x+2) + \frac{2x^2+3x+1}{x^3-2x})A(x) + ((x+1) + \frac{4x-1}{x^3-2x})B(x)=1$$

Now, $$1=0*A(x)+\frac{1}{B(x)}B(x)$$

Now, we can make some system, what do you think?

 

[Physicsissuef]

tiny-tim what do you think?

 

[tiny-tim]

The answer …

Hi Theofilius and Physicsissuef!

The solution is:

$$A(x)\,=\,(3x^3\,+\,3x^2\,-\,7x\,+\,2)\,;\,B(x)\,=\,(-3x^3\,-6x^2\,+\,x\,+\,2)\,.$$ ​

But the way I go it is fairly horrible , so I'm going to carry on thinking about it - I'm sure there must be a simpler way!

Theofilius, what did they teach you in class about remainders? Maybe there's a clue in there somwehere …

 

[Physicsissuef]

Hi Theofilius and Physicsissuef!

The solution is:

$$A(x)\,=\,(3x^3\,+\,3x^2\,-\,7x\,+\,2)\,;\,B(x)\,=\,(-3x^3\,-6x^2\,+\,x\,+\,2)\,.$$ ​

But the way I go it is fairly horrible , so I'm going to carry on thinking about it - I'm sure there must be a simpler way!

Theofilius, what did they teach you in class about remainders? Maybe there's a clue in there somwehere …

What way you were using? Is my way, correct?

 

[tiny-tim]

What way you were using? Is my way, correct?

erm … if you must know …
I put P = 1/2(A+B), Q = 1/2(A-B),
then Q = xR, then P + R/2 = U, 3P/2 + R = V, which gave me:

$$x^2(x^2\,-\,1)U\,+\,(x^2\,+\,1)V\,=\,x^2\,-\,2\,.$$ ​

Sorry, but I didn't understand your way … how does it go on?

 

[Physicsissuef]

First, I divide by [itex]x^3-2x[/tex]
and then
$$((x+2) + \frac{2x^2+3x+1}{x^3-2x})A(x) + ((x+1) + \frac{4x-1}{x^3-2x})B(x)=1$$

$$((x+2) + \frac{2x^2+3x+1}{x^3-2x})A(x) + ((x+1) + \frac{4x-1}{x^3-2x})B(x)=0*A(x)+\frac{1}{B(x)}B(x)$$

 

[tiny-tim]

Hi Physicsissuef!

I understand how you get from the first line to the second line … but not why.

Where would you go from there?

 

[Physicsissuef]

$$\frac{2x^2+3x+1}{x^3-2x}=0$$

$$\frac{4x-1}{x^3-2x}=\frac{1}{B}$$

I could find B, probably out of here.

 

[Physicsissuef]

Or It is better to write:
$$\frac{2x^2+3x+1}{x^3-2x}=\frac{2}{A}$$

$$\frac{4x-1}{x^3-2x}= - \frac{1}{B}$$

 

[tiny-tim]

Physicsissuef, A and B are polynomials (cubic ones, as it turns out).

You can't treat them as if they were numbers and expect them to make valid fractions!

I really can't see this working …

 

[Physicsissuef]

If u substitute the given polynoms (which are A and B), you will see that it works.
Anyway, can you explain me, please, how did you find the polynom? I can't understand anything from your post.

 

[tiny-tim]

Physicsissuef, I have explained it!

The proof is far too long for me to type out in LaTeX.

But if you follow my substitutions (P and Q for A and B, and so on), slowly and carefully, you'll find that it works!

oh, and the solution to

$$x^2(x^2\,-\,1)U\,+\,(x^2\,+\,1)V\,=\,x^2\,-\,2\,.$$​

is, of course, U = -3/2, V = 3$$x^2$$/2 - 2.

 

[Physicsissuef]

Are you talking about the polynom that Theofilius had given, or this is yours polynom?

 

[tiny-tim]

Are you talking about the polynom that Theofilius had given, or this is yours polynom?

Theofilius' of course - if you put my A(x) and B(x) into his original question, and multiply it out, you'll find that it does work (actually, I assumed you'd already done that!)

 

[Physicsissuef]

Can you please start from beginning and write the problem without using latex. Why you used 1/A-B and 1/A+B, and so on... Please!

 

[tiny-tim]

Can you please start from beginning and write the problem without using latex. Why you used 1/A-B and 1/A+B, and so on... Please!

Oh, sorry, when I put P = 1/2(A+B), Q = 1/2(A-B), I meant P = (A+B)/2, Q = (A-B)/2.

No, Physicsissuef, you do it.

You won't understand anything just by seeing my proof.

Start by writing P = (A+B)/2, Q = (A-B)/2 into the original equation.

You should get an equation which makes it clear that Q must be divisible by x. So Put Q = xR.

… and so on …

If you have any particular difficulty, you can get back to me on it.

However, it's not a very sensible proof anyway, and I really don't recommend you trying to understand it …

 

[Physicsissuef]

Where I substitute P and Q exactly and why you get A+B/2 and A-B/2, why not some other numbers?