Solving polynomial equation using induction

[thereddevils]

Homework Statement

If f(x)=(x+1)p(x) where f(x)=x^{2n}+2nx+2n-1, what is p(x)?

Answer given: x^{2n-1}-x^{2n-2}+...-x^2+x+2n-1

Homework Equations

The Attempt at a Solution

I tried the long division and managed to some terms correct. Is there any other methods of finding this? Perhaps induction?

 

[Mentallic]

I barely ever use long division, because it is easy to find ways around it which I prefer. I can't see why it wouldn't work though.

x^{2n}+2nx+2n-1=(x-1)p(x)

So how about we equate coefficients?

Say, if we were given that x^3-x^2+x-1=(x-1)f(x) then we could let f(x)=a_1x^2+a_2x+a_3 where a_n is just some constant, then we would expand the right hand side as such, a_1x^3+a_2x^2+a_3x-a_1x^2-a_2x-a_3=a_1x^3+(a_2-a_1)x^2+(a_3-a_2)x-a_3

So now we can equate coefficients, since x^3-x^2+x-1=a_1x^3+(a_2-a_1)x^2+(a_3-a_2)x-a_3

Then obviously,

a_1=1
a_2-a_1=-1
a_3-a_2=1
-a_3=-1

Now just do the same for your problem. You can of course skip obvious steps by quickly realizing that the coefficient of the x³ term is 1, so the other factor a_1=1 and also the constant is equal to 1, so a_3=-1.

 

[thereddevils]

I barely ever use long division, because it is easy to find ways around it which I prefer. I can't see why it wouldn't work though.

x^{2n}+2nx+2n-1=(x-1)p(x)

So how about we equate coefficients?

Say, if we were given that x^3-x^2+x-1=(x-1)f(x) then we could let f(x)=a_1x^2+a_2x+a_3 where a_n is just some constant, then we would expand the right hand side as such, a_1x^3+a_2x^2+a_3x-a_1x^2-a_2x-a_3=a_1x^3+(a_2-a_1)x^2+(a_3-a_2)x-a_3

So now we can equate coefficients, since x^3-x^2+x-1=a_1x^3+(a_2-a_1)x^2+(a_3-a_2)x-a_3

Then obviously,

a_1=1
a_2-a_1=-1
a_3-a_2=1
-a_3=-1

Now just do the same for your problem. You can of course skip obvious steps by quickly realizing that the coefficient of the x³ term is 1, so the other factor a_1=1 and also the constant is equal to 1, so a_3=-1.

Thank you. I should have thought of this.