prove-that-x-2-divide-x-1-n-nx-1-nintegr-

Question Number 31494 by abdo imad last updated on 09/Mar/18

p r o v e t h a t x^{2} d i v i d e {(x + 1)}^{\underset{}{n}} - n x - 1 . n i n t e g r .

Commented by Rasheed.Sindhi last updated on 10/Mar/18

{(x + 1)}^{n} - nx - 1

x^{n} + {nx}^{n - 1} + \dots (\begin{matrix} n \\ n - 2 \end{matrix}) x^{2} + nx + 1 - nx - 1

x^{n} + {nx}^{n - 1} + \dots (\begin{matrix} n \\ n - 2 \end{matrix}) x^{2}

x^{2} (x^{n - 2} + {nx}^{n - 3} + \dots (\begin{matrix} n \\ n - 2 \end{matrix}))

Hence x^{2} divides {(x + 1)}^{n} - nx - 1

Commented by abdo imad last updated on 09/Mar/18

l e t r e m e m b e r t h i s t h e o r e m a r o o t s o f p (x) w i t h o r d r m \Leftrightarrow

p (a) = p^{^{'}} (a) = \dots = p^{m - 1} (a) = 0 a n d p^{(m)} (a) \neq 0 a n d i f

p (a) = p^{^{'}} (a) = \dots = p^{(m - 1)} (a) = 0 \Rightarrow {(x - a)}^{m} d i v i d e p (x)

h e r e w e h a v e p (x) = {(x + 1)}^{n} - n x - 1 \Rightarrow p (0) = 0

p^{^{'}} (x) = n {(x + 1)}^{n - 1} - n \Rightarrow p^{^{'}} (0) = n - n = 0 \Rightarrow x^{2} d i v i d e p (x)

p l u s t h a t w e h a v e p^{^{″}} (x) = n (n - 1) {(x + 1)}^{n - 2} \Rightarrow p^{^{″}} (p 0) = n (n - 1)

p^{^{″}} (0) \neq 0 \Rightarrow 0 i s r o o t o f p (x) a r o r d r e 2 .

Commented by abdo imad last updated on 09/Mar/18

y o u r a n s w e r i s a l s o c o r r e c t s i r r a c h i d \dots