# Double root (Differentiation)

1. Nov 18, 2012

### mtayab1994

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

$$p(x)=vx^{n+1}+ux^{n}+1$$

2. Relevant equations

1) Find u and v so that 1 is a double root for p.

2) Conclude the quotient of p(x) over (x+1)^2.

3) For n=4 find u and v and find the quotient of p(x) over (x-1)^2.

3. The attempt at a solution

Can someone just tell me how to start this please. I really have no idea on how to start this off. Thanks in advance.

Last edited: Nov 18, 2012
2. Nov 18, 2012

### Fightfish

Edit: Sorry something else came to my mind.

1) If 1 is double root, then the graph of p(x) just touches the x-axis but does not cross it at x=1. This means that it must be a stationary point.

Last edited: Nov 18, 2012
3. Nov 18, 2012

### HallsofIvy

Staff Emeritus
Yes, that's correct. At x= 1, p(1)= u+ v+ 1. In order that 1 be zero of p, we must have u+ v+ 1= 0. The derivative of p is p'(x)= (n+1)vx^n+ nux^{n-1}. In order that 1 be a double zero of p we must have p'(1)= (n+1)v+ nu= 0. That gives you two equations to solve for u and v.

4. Nov 18, 2012

### mtayab1994

After solving the system of equations that contained u+v+1=0 and (n+1)v+nu=0 I found that:

1)v=n and u=-n-1

2) The quotient of p when divided by (x-1)^2 will be $$nx^{n-1}+(n-1)x^{n-2}+(n-2)x^{n-3}+...+1$$

3)When we take n to be 4 the quotient comes out to be 4x^3+3x^2+2x+1.

Is everything correct??

Last edited: Nov 18, 2012