# Homework Help: Polynomial division homework

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1. Nov 18, 2015

### stungheld

1. The problem statement, all variables and given/known data
How many pairs of solutions make x^4 + px^2 + q = 0 divisable by x^2 + px + q = 0

2. Relevant equations
x1 + x2 = -p
x1*x2= q

3. The attempt at a solution
I tried making z = x^2 and replacing but got nowhere. I figure 0,1,-1 are 3 numbers that fit but im not sure whats being asked also. How many pairs of x1 and x2 make the polynomial 1 divisible by polynomial 2. How to start solving this?

2. Nov 18, 2015

### HallsofIvy

The second polynomial can be written as $(x- a)(x- b)= x^2- (a+ b)x+ ab$ with a+ b= -p and ab= q. It divides the first if and only if the first polynomial has those same factors: is of the form $(x- a)(x- b)(x- c)(x- d)= x^4- (a+ b+ c+ d)x^3+ (ab+ ac+ ad+ bc+ bd+ cd)x^2- (abc+ acd+ bcd)x+ abcd$ with a+ b+ c+ d= 0, ab+ ac+ ad+ bc+ bd+ cd= p, abc+ acd+ bcd= 0, and abcd= q. See what you can make of those 6 equations.

3. Nov 18, 2015

### Samy_A

Just a guess:

Maybe the question is:
For how many pairs (p,q) is the polynomial $x^4+px^2+q$ divisible by the polynomial $x^2+px+q$?
Where divisible means that the quotient is a polynomial too.

Admittedly, that doesn't explain the "=0" in the question, so I'm far for certain.

EDIT: didn't see the post above before I posted, so just ignore this.

Last edited: Nov 18, 2015
4. Nov 18, 2015

### PeroK

This is a good question and will stretch your understanding of polynomials. First, I assume that you've added the "=0". It's important to know the difference between a polynomial and a polynomial equation. In this case you are dealing with polynomials. You are not trying to find the values of $x$ for which the polynomial is 0.

I would rephrase the question as:

Find $p$ and $q$ such that $x^2 + px + q$ divides $x^4 + px^2 + q$

Can you see the first step?

5. Nov 19, 2015

### stungheld

It is true what you said. I added = 0. It shouldnt be there. I dont seem to have any idea on how to start, maybe you could help me with some subtle hint without giving me the step?

6. Nov 19, 2015

### PeroK

In general, the definition of $A$ divides $B$ is that there exists some $C$ such that $B = AC$

7. Nov 19, 2015

### ehild

Do what you wrote in the title: Polynomial Long Division. Divide x4+px2+q with x2+px+q, and find the values of p and q so as the remainder is zero.

8. Nov 21, 2015

### stungheld

I havent done long division before so i checked it out online and tried to do it.Could you check this?

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9. Nov 21, 2015

### ehild

You have sign errors at the last stage.

10. Nov 21, 2015

### stungheld

I dont see it

11. Nov 21, 2015

### SammyS

Staff Emeritus
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Those are the last three lines of the image you posted.

You need to subtract the part enclosed in the red parentheses.

12. Nov 22, 2015

### stungheld

(p^2 + p - q)(- px - q) + q(px + 1)= 0
This must be true for some p and q so that there is no remainder. But how do i find those values? There are three variables here.

13. Nov 22, 2015

### ehild

They must be true for any value of x. Collect the x terms and the constant terms separately, in form A(p,q)x +B(p,q)≡0. Both A and B are equal to zero. So you have two equations with the parameters p,q.

14. Nov 22, 2015

### stungheld

Well i get x(-p^3 - p^2 + 2pq) + (-qp^2 - qp + q^2 + q) = 0
So like you said A and B are 0. I set them to 0 and solved for p but got some cubic equation so i supposed three solutions there. But i cant solve for q.

15. Nov 22, 2015

### vela

Staff Emeritus
I suggest you avoid polynomial division altogether and use HallsofIvy's suggestion back in post #2. He made a slight error in the coefficient of the linear term, however. It should be (abc + abd + acd + bcd).

Consider the cases q = 0 and $q \ne 0$ separately.

16. Nov 22, 2015

### stungheld

That seems more complicated. Any chance we could try resolving this polynomial division? How to check the solutions after the last part of the division? I separated the cases in the x(-p^3 - p^2 + 2pq) + (-qp^2 - qp + q^2 + q) = 0. I tried to eliminate one variable by substituting p for q but gotten -2p^3 + 2p^2 - p = 0, and when i try the other way around i get something like p^2 + p = q - 1

17. Nov 22, 2015

### ehild

You can factorize both equations, getting p(-p^2-p+2q)=0 and q(-p^2-p+q+1)=0. What solutions are possible?

Last edited: Nov 22, 2015
18. Nov 22, 2015

### stungheld

p and q = 0
q = 0 and p either of two solutions
p = 0 and q = -1
p = q = 1
That makes it 5. Is that it?

19. Nov 22, 2015

### ehild

Make a table for the possible values of p and q.
It looks that you counted p=0 q=0 twice, and one solution is still missing. What are the solutions if neither p nor q are zero?

Last edited: Nov 22, 2015
20. Nov 22, 2015

### stungheld

How did i count p and q to be 0 twice? I counted both to be 0, two cases were in the first q is 0 and p p1 or p2 ( making the first expression 0) and the only case were p and q are non zero but still make zero are p = q = 1. I see no other way

21. Nov 22, 2015

### ehild

So you have so far
p=0, q=0
p=0, q=-1
p? , q=0
plus the solutions of
p^2+p-2q=0 ,
p^2+p+-q-1=0
one of them is p=1, q=1, what is the other one?

22. Nov 22, 2015

### stungheld

If q = 0 p = -1 and for the last one its p = -2 and q = 1
5 solutions would then be the answer

23. Nov 22, 2015

Perfect!