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?

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)