Question

Show that (x-1) is a factor of x^n-1

please give correct answer!

Answer 1

$f(x) = x {}^{n} - 1$

$g(x) = x - 1$

$x - 1 = 0$

$put \: x = 1 \: \: \: \ in \: f(x)$

$(1) {}^{n} - 1$


No matter what value of n it is always be 1

so 1 - 1= 0

Hence (x - 1) is a factor of f(x)



Thanks

Answer 2

Heya!!
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here's your answer!!
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here is your answer which u are searching for.

let us proof it by induction

let n = 1

${x}^{0} - 1 = 1 - 1 = 0$

let us assume that the statement is true for some value k

As it is true for k it must be true for
k+1

so, our first assumption is that x-1 is a factor of
${x}^{k + 1}$

$let \: us \: divide \: x - 1 \: into \: {x}^{k + 1} - 1 \: it \: end \\ \: up \: with \: {x}^{k - 1}$

Now, divide x-1 into that......

$hence \: (x - 1) \: is \: a \: factor \: of \: {x}^{n - 1}$

Glad help you
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it helps you
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thank you
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@vaibhavhoax
# born with attitude
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