Proof that polynome $P(x)$ is divisible by $Q(x)$

Question

Prove that the polynome $$P(x)=nx^{n+1}-(1+n\lambda)x^n+(\lambda-1)(x^{n-1}+...+x)+\lambda$$ is divisible by: $$Q(x)=x^2-(\lambda+1)x+\lambda$$ where $\lambda\in\Bbb{C}$

So first I found the roots of $Q(x)$, assuming if $P$ is divisible by $Q$, the remainder must be $0$.

$$x^2-(\lambda+1)x+\lambda=0\iff x_{1/2}={\lambda+1\pm\sqrt{(\lambda+1)^2-4\lambda}\over2}$$

from here, since $\lambda\in\Bbb{C}$, I can say:

$x_1=\lambda\\x_2=1$

now I thought I should find $P(\lambda)$. After a long simplification I got that:

$$P(\lambda)=\lambda^{n+2}-2\lambda^{n+1}-\lambda^3+2\lambda^2$$

at this point I'm lost. This would equal to zero if $\lambda=0$ but that doesn't prove me anything.

Answer 1

Alternatively, it is easy to check that $$P(x)=Q(x)(nx^{n-1}+(n-1)x^{n-2}+\dots+3z^2+2x+1)$$ That still works for $\lambda=1$ whereas the approach of checking that $1,\lambda$ are roots of $P(x)$ requires a continuity argument (or checking the derivative) for $\lambda=1$.

Answer 2

You have found the roots.

It is straightforward to see that \begin{align*} P(1) =& n-(1+n \lambda) + (\lambda -1)(n-1) + \lambda \\ =& n-1-n \lambda + n\lambda-n-\lambda+1+\lambda \\ =& 0\\ \\ P(\lambda) =& n \lambda^{n+1} - (1+n\lambda)\lambda^n + (\lambda-1)(\lambda^{n-1} + \ldots + \lambda) + \lambda \\ =& n\lambda^{n+1} - \lambda^n-n\lambda^{n+1}+\lambda^n-\lambda+\lambda \\ =& 0 \end{align*} Hence, both $x-1$ and $x-\lambda$ are divisors of $P$, which means $Q(x) = (x-1)(x-\lambda)$ must also be one.