Prove that an expression is divisible by a polynomial Question: Show that the polynomial $f(x)=(x+1)^{2n} +(x+2)^n  -  1$
is divisible by $g(x) = x^2+3x+2$, where $n$ is an integer.
I have tried to use mathematical induction. The basis case wasn't that difficult, but when it comes to the inductive step itself, I got a bit confused.
Is it possible to prove this by mathematical induction, and is binominal expansion required at that step?
 A: Putting the zeros of $x^2+3x+2=0$ i.e., $-1,-2$ one by one, in $f(x)=(x+1)^{2n}+(x+2)^n-1$ 
we get, $f(-1)=(-1+1)^{2n}+(-1+2)^n-1=0$
and $f(-2)=(-2+1)^{2n}+(-2+2)^n-1=0$ 
(i)So, using Remainder Theorem, 
$(x+1)\mid f(x)$ and $(x+2)\mid f(x)\implies lcm(x+1,x+2)\mid f(x)$
(ii) Alternatively, 
$\frac{(x+1)^{2n}+(x+2)^n-1}{x+1}=(x+1)^{2n-1}+\frac{(x+2)^n-1}{x+1}$
Now $x+1(=x+2-1)\mid \{(x+2)^{2n-1}-1\} $ as $(a-b)\mid (a^n-b^n)--->(1)$
So, $(x+1)\mid f(x).$
$\frac{(x+1)^{2n}+(x+2)^n-1}{x+2}=\frac{\{(x+1)^2\}^n-1}{x+2}+(x+2)^{n-1}$
Using $(1),\{(x+1)^2\}^n-1$ is divisible by $(x+1)^2-1=x^2+2x=x(x+2)$
So, $(x+2)\mid f(x).$
$lcm(x+1,x+2)=(x+1)(x+2)=x^2+3x+2$ (prove this)
A: Hint $\rm\: mod\ g\!:\,\ e_i^2\! = e_i\Rightarrow\: e_1^n + e_2^n =\, e_1\! + e_2.\ $ Google (orthogonal) idempotents to learn more.
A: Warning: Non-Induction Proof Put $z = x+1$ and use $z$ rather than $x$ but everything else is the same. So let's define $$f_n(z)=z^{2n} + (z+1)^n - 1.$$ Obviously $z|f_n(z)$ and $z$ is coprime to $z+1$ so we just need to show that $z+1|f_n(z)$ but that is equivalent to $-1$ being a root of the polynomial, which we verify it is by evaluation.
A: I don't have an answer for you because I didn't exercise this stuff a long time but I did find a reference that might assist you.
Please try : http://www.enotes.com/math/q-and-a/determine-m-polynomial-p-divisible-by-x-3-p-x-3-mx-188671/
Using the remainder theorem "http://en.wikipedia.org/wiki/Polynomial_remainder_theorem" 
Hope It helps a little bit,
Guy
