# Find polynoms, with as least as power possible

1. Mar 8, 2008

### Theofilius

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

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

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

2. Relevant equations

3. The attempt at a solution

2. Mar 9, 2008

3. Mar 9, 2008

### 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!

4. Mar 9, 2008

### Theofilius

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

5. Mar 9, 2008

### HallsofIvy

Staff Emeritus
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.

6. Mar 9, 2008

### Theofilius

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

7. Mar 9, 2008

### tiny-tim

(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.

8. Mar 9, 2008

### Theofilius

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

9. Mar 9, 2008

### tiny-tim

: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?

10. Mar 9, 2008

### 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)?

11. Mar 9, 2008

### tiny-tim

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, …

12. Mar 9, 2008

### Theofilius

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

13. Mar 10, 2008

### 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.

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14. Mar 10, 2008

### Theofilius

There are 1000 of combinations, isnt any simpler way of doing this job?

15. Mar 10, 2008

### 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) …

16. Mar 10, 2008

### Physicsissuef

it has no solution

17. Mar 11, 2008

### 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.

18. Mar 12, 2008

### 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.

19. Mar 12, 2008

### 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?

Last edited: Mar 13, 2008
20. Mar 13, 2008

### Physicsissuef

tiny-tim what do you think?