Question

show that x-1 is a factor of x power20 - 1 andxpower21- 1​

Answer 1

Answer:

Given,

x-1=0

x=1

p(x)=x^20-1 & f(x)=x^21-1

=> p(x)

(1)^20-1

1-1

0

=> f(x)

(1)^21-1

1-1

0

Therefore x-1 is factor of x^20-1 and x^21-1.

Answer 2

Answer:

Hello dear user

Hello dear user Here is your answer..

$given \: p(x) = {x}^{20} + 1 \\ and \: f(x) = {x}^{21} - 1 \\ we \: have \: to \: show \: that \: x - 1 \: is \: factor \: of \: p(x) \: and \: f(x) \\ x - 1 = 0 \\ x = 1 \\ putting \: x = 1 \: in \: both \: p(x) \: and \: g(x) \\ p(x) = {x}^{20} - 1 \\ p(1) = {1}^{20} - 1 \\ p(x) = 0 \\ now \: in \: f(x) = {x}^{21} - 1 \\ f(1) = {1}^{21} - 1 \\ f(1) = 0 \\ so \: both \: p(x) \: and \: f(x) \: come \: zero \: \\ so \: (x - 1) \: is \: a \: factor \: of \: both \: p(x) \: and \: f(x)$

Hope it is clear to you.