Prove that $(n-2m)x^n-nx^{n-m}+nx^m-(n-2m)$ is divisble by $(x-1)^3$

Question

Prove that $(n-2m)x^n-nx^{n-m}+nx^m-(n-2m)$ is divisble by $(x-1)^3$, where $n,m \in {\displaystyle \mathbb {N}}$

I have no idea how to start.

Thanks in Advance.

Answer 1

If a polynomial can be written as:

$p(x)=(x-1)^{3}g(x)$, (where $g(x)$ is a polynomial) , then,

$p'(x)=3(x-1)^{2}g(x)-(x-1)^{3}g'(x)$

→$p'(x)=(x-1)^{2}[somepolynomial]$

Similarly,

$p''(x)=(x-1)[somepolynomial]$

hence we can say that, whenever a polynomial $p(x)$ is divisible by $(x-1)^{3}$, then

1) $p(1)=0$ (i.e. $(x-1)|p(x)$)

2) $p'(1)=0$ (i.e. $(x-1)|p'(x)$)

3) $p''(1)=0$ (i.e. $(x-1)|p''(x)$)

(the converse can also be proved)

In this problem, we can observe,

1)$p(1)=(n-2m)-n+n-(n-2m)=0$

2)$p'(x)=n(n-2m)x^{n-1}-n(n-m)x^{n-m-1}+nmx^{m-1}$

→$p'(1)=n(n-2m)-n(n-m)+nm=0$

3)$p''(1)=n(n-2m)(n-1)-n(n-m)(n-m-1)+nm(m-1)=0$

Hence $(x-1)^{3}|p(x)$

Answer 2

Hint: Divide your polynomial by $${x}^{3}-3\,{x}^{2}+3\,x-1$$

Answer 3

hint...first show $f(1)=0$ then differentiate twice and show that $(x-1)$ is a factor of the resulting polynomials i.e. show $f'(1)=0$ and $f''(1)=0$

Answer 4

Well, one can prove that iff $(x-1)^3$ divides a polynomial, then $(x-1)^2$ and $(x-1)$ should divide the first and second derivatives of said polynomial, respectively. So then, what should you do? If you need further aid, update first with your progress.