how do you factor this polynomial?

x^(2k+1) - y^(2k+1)

my teacher's answer is:

[x^2 (x^(2k-1) + y^(2k-1))] + [(y^2 - x^2)y^(2k-1)]

what is the reasoning involved to factor something as complicated as this?

Printable View

- Mar 3rd 2009, 03:08 PMoblixpscomplicated factoring problem
how do you factor this polynomial?

x^(2k+1) - y^(2k+1)

my teacher's answer is:

[x^2 (x^(2k-1) + y^(2k-1))] + [(y^2 - x^2)y^(2k-1)]

what is the reasoning involved to factor something as complicated as this? - Mar 3rd 2009, 04:20 PMAbu-Khalil
$\displaystyle x^{2k+1}-y^{2k+1}=x^{2k+1}+x^2y^{2k-1}-x^2y^{2k-1}-y^{2k+1}=x^2\left(x^{2k-1}-y^{2k-1}\right)-\left(x^2-y^2\right)y^{2k-1}$

It's easily to see in this way when you are using induction to prove divisibility. - Mar 3rd 2009, 07:57 PMSoroban
Hello, oblixps!

Quote:

How do you factor this polynomial?

. . $\displaystyle x^{2k+1} - y^{2k+1}$

my teacher's answer is: .$\displaystyle \bigg[x^2 \left(x^{2k-1} + y^{2k-1}\right)\bigg] \:{\color{red}+}\: \bigg[\left(y^2 - x^2\right)y^{2k-1}\bigg]$

Wrong! . . . That is not factored!

The difference of two odd powers can always be factored.

$\displaystyle x^{2k+1} - y^{2k+1} \;=\;\bigg(x-y\bigg)\,\bigg(x^{2k} + x^{2k-1}y + x^{2k-2}y^2 + x^{2k-3}y^3 + \hdots + x^2y^{2k-2}$ $\displaystyle + xy^{2k-1} + y^{2k} \bigg)$