How do I demonstrate that a polynomial of degree $2$ divides one of degree $n$?

Let $f$ and $g$ the polynomials

$$f(x) = (x+1)^{2n-1}+(-1)^n(x+2)^{n+1}\qquad\text{and}\qquad g(x) = x^2 + 3x + 3$$

How do I demonstrate that $g$ divides $f$? I tried finding the roots of $g$ then testing them for $f$, it didn't work ...

I also wrote g as $(x+1)^2 + x + 2$ but I couldn't get anywhere.

• Your second approach looks promising... Is it possible to separate even and odd cases of $n$? Have you worked with small $n$, or tried induction? – abiessu Mar 28 '16 at 13:26
• Another approach might be to rewrite using $y=x+1$ and see what happens. – abiessu Mar 28 '16 at 13:31
• I did this $(x+1)^2 + x + 2 = 0 <=> (x+1)^2 = (-1)(x+2) |^n <=> (x+1)^{2n} = (-1)^n(x+2)^n$ but i don't know now what to do – oren revenge Mar 28 '16 at 13:32

Let $r$ be any of the two roots of $g$; since $g(r) = (r+1) (r+2) + 1$, obviously $r+1 \ne 0$ and we get $r+2 = -\frac 1 {r+1}$. To have $g \mid f$ we should have $f(r) = 0$, so let us compute $f(r)$.

Since

$$f(r) = (r+1)^{2n-1} + (-1)^n (r+2)^{n+1} = (r+1)^{2n-1} + (-1)^n \left(- \frac 1 {r+1} \right)^{n+1} = \\ (r+1)^{2n-1} - \frac 1 {(r+1)^{n+1}} = \frac {(r+1)^{3n} - 1} {(r+1)^{n+1}} ,$$

let us finally show that $(r+1)^{3n} - 1 = 0$. To do this, notice that

$$(r+1)^3 = r^3 + 3r^2 + 3r + 1 = r g(r) + 1 = 1 ,$$

whence it follows that $(r+1)^{3n} - 1= \left((r+1)^3 \right)^n - 1 = 1^n - 1 = 0$, as desired.

This shows that indeed $f(r) = 0$ for each of the two roots $r$ of $g$, so indeed $g \mid f$.

For $f_n(x) = (x+1)^{2n-1}+(-1)^n(x+2)^{n+1}$ with $g(x) = x^2+3x+3=(x+1)^2+x+2$, first consider $n=1,2$:

$$f_1(x) = x+1-x^2-4x-4 = -x^2-3x-3\\ f_2(x) = (x+1)^3+(x+2)^3=(2x+3)\left((x+1)^2-(x+1)(x+2)+(x+2)^2\right)\\ =(2x+3)(-(x+1)+(x+2)^2)=(2x+3)(x^2+3x+3)$$

So we have that $g(x)\mid f_1(x),\ g(x)\mid f_2(x)$. It appears that a useful way to approach the question is to show that we end up with something either of the form $(x+1)^2+(x+2)$ or $(x+2)^2-(x+1)$. To this end, with $y=x+1\to g(y)=y^2+y+1$, consider that we have

$$f_n(y) = y\left(y^2\right)^{n-1}+(-1)^n(y+1)^2(y+1)^{n-1}\\ =(-1)^n(y^2+y+1)(y+1)^{n-1}+y\left((y^2)^{n-1}+(-1)^n(y+1)^{n-1}\right)$$

For even $n$, $(y^2)+(y+1)$ is always a factor of $(y^2)^{n-1}+(-1)^n(y+1)^{n-1}$ (just as $a+b\mid a^{2n-1}+b^{2n-1}$). Now the only remaining issue is to determine whether $(y^2)+(y+1)$ is a factor of $f_n(x)$ for odd $n$. But we see that for odd $n=2m+1$, $(y^2)^{n-1}+(-1)^n(y+1)^{n-1}$ is a difference of squares, and therefore factors as

$$\left((y^2)^m-(y+1)^m\right)\left((y^2)^m+(y+1)^m\right)$$

If $m$ is odd, we are done; if $m$ is even, we can split the left-hand term as a difference of squares again; we can repeat as needed until we have a sum with odd outer exponents. Therefore $g(y)\mid f_n(y)$ and thus $g(x)\mid f_n(x)$.