The discussion focuses on solving the polynomial equation f(x) = (x+1)p(x) where f(x) = x^{2n} + 2nx + 2n - 1. The solution provided identifies p(x) as x^{2n-1} - x^{2n-2} + ... - x^2 + x + 2n - 1. Participants explore methods such as polynomial long division and coefficient comparison to derive p(x), emphasizing the effectiveness of equating coefficients in polynomial identities.
Students and educators in mathematics, particularly those focusing on algebra and polynomial equations, as well as anyone interested in enhancing their problem-solving skills in mathematical proofs.
Mentallic said: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.
[tex]x^{2n}+2nx+2n-1=(x-1)p(x)[/tex]
So how about we equate coefficients?
Say, if we were given that [tex]x^3-x^2+x-1=(x-1)f(x)[/tex] then we could let [tex]f(x)=a_1x^2+a_2x+a_3[/tex] where [tex]a_n[/tex] is just some constant, then we would expand the right hand side as such, [tex]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[/tex]
So now we can equate coefficients, since [tex]x^3-x^2+x-1=a_1x^3+(a_2-a_1)x^2+(a_3-a_2)x-a_3[/tex]
Then obviously,
[tex]a_1=1[/tex]
[tex]a_2-a_1=-1[/tex]
[tex]a_3-a_2=1[/tex]
[tex]-a_3=-1[/tex]
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 [tex]a_1=1[/tex] and also the constant is equal to 1, so [tex]a_3=-1[/tex].