Prove that (x-1) is a factor of x^17-1.please friends urgent

1.	Prove that (x-1) is a factor of x^17-1.please friends urgent
Answer» Note : ★ Remainder theorem : If a polynomial p(x) is DIVIDED by (x - c) , then the remainder OBTAINED is given as R = p(c) . ★ FACTOR theorem : If the remainder obtained on dividing a polynomial p(x) by (x - c) is zero , ie. if R = p(c) = 0 , then (x - c) is a factor of the polynomial p(x) . If (x - c) is a factor of the polynomial p(x) , then the remainder obtained on dividing the polynomial p(x) by (x - c) is zero , ie. R = p(c) = 0 . Solution : Given polynomial : x^(17) - 1 To prove : (x - 1) is a factor of x^(17) - 1 Proof : Let the given polynomial be ; p(x) = x^(17) - 1 Also , If x - 1 = 0 , then x = 1 Now , By remainder theorem , the remainder on diving the given polynomial p(x) by (x - 1) will be given as ; => R = p(1) => R = 1^(17) - 1 => R = 1 - 1 => R = 0 Since , The remainder obtained on dividing [x^(17) - 1] by (x - 1) is zero , thus (x - 1) is a factor of [x^(17) - 1] . Hence proved .

Answer»

Note :

★ Remainder theorem : If a polynomial p(x) is DIVIDED by (x - c) , then the remainder OBTAINED is given as R = p(c) .

★ FACTOR theorem :

If the remainder obtained on dividing a polynomial p(x) by (x - c) is zero , ie. if R = p(c) = 0 , then (x - c) is a factor of the polynomial p(x) .

If (x - c) is a factor of the polynomial p(x) , then the remainder obtained on dividing the polynomial p(x) by (x - c) is zero , ie. R = p(c) = 0 .

Let the given polynomial be ;

p(x) = x^(17) - 1

Also ,

If x - 1 = 0 , then x = 1

Now ,

By remainder theorem , the remainder on diving the given polynomial p(x) by (x - 1) will be given as ;

=> R = p(1)

=> R = 1^(17) - 1

=> R = 1 - 1

=> R = 0

Since ,

The remainder obtained on dividing [x^(17) - 1] by (x - 1) is zero , thus (x - 1) is a factor of [x^(17) - 1] .