Finding Solutions for Polynomial Division: Where to Begin?

[stungheld]

Homework Statement

How many pairs of solutions make x^4 + px^2 + q = 0 divisable by x^2 + px + q = 0

Homework Equations

x1 + x2 = -p
x1*x2= q[/B]

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 I am not sure what's being asked also. How many pairs of x1 and x2 make the polynomial 1 divisible by polynomial 2. How to start solving this?

 

[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.

 

[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.

 

[PeroK]

Homework Statement

How many pairs of solutions make x^4 + px^2 + q = 0 divisable by x^2 + px + q = 0

Homework Equations

x1 + x2 = -p
x1*x2= q[/B]

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 I am not sure what's being asked also. How many pairs of x1 and x2 make the polynomial 1 divisible by polynomial 2. How to start solving this?

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?

 

[stungheld]

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?

It is true what you said. I added = 0. It shouldn't be there. I don't seem to have any idea on how to start, maybe you could help me with some subtle hint without giving me the step?

 

[PeroK]

It is true what you said. I added = 0. It shouldn't be there. I don't seem to have any idea on how to start, maybe you could help me with some subtle hint without giving me the step?

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

 

[ehild]

I don't seem to have any idea on how to start, maybe you could help me with some subtle hint without giving me the step?

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.

 

[stungheld]

I haven't done long division before so i checked it out online and tried to do it.Could you check this?

 

[ehild]

I haven't done long division before so i checked it out online and tried to do it.Could you check this?

You have sign errors at the last stage.

 

[stungheld]

I don't see it

 

[SammyS]

I haven't done long division before so i checked it out online and tried to do it.Could you check this?

(Your reply to ehild:

I don't see it

)

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

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

 

[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.

 

[ehild]

(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.

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.

 

[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 can't solve for q.

 

[vela]

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.

 

[stungheld]

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.

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

 

[ehild]

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 can't solve for q.

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

 

[stungheld]

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

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?

 

[ehild]

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?

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?

 

[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

 

[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?

 

[stungheld]

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

 

[ehild]

Perfect!