When x^31 + x^17 + 1 is divided by x-1, what is the remainder?

Question

Please show work so I'll know how to do it. Thanks!

Answer 1

f( x) = x^31 + x^17 + 1 is a plynomial equation

by remainder theorem (x-a) leaves remainder f(a)(link below)

so remainder when devide by x-1 is

f(1) = 1 + 1 + 1 = 3

Answer 2

p(x)/(x-a) = q(x) with r(x) where q is the quotient and r is the remainder; or, p(x) = (x – a)q(x) + r(x).

Notice what happens if x=a

p(a) = (a-a)q(x) + r(x)

p(a) = (0)q(x) + r(x)

p(a) = r(x)

Thus, the remainder can be found by evaluating p(a)

p(a) = r(x) = 1^31 + 1^17 + 1

r(x) = 1 + 1 +1

r(x) = 3

Answer: 3

Answer 3

Use the remainder theorem for this.

f(x) = x^31 + x^17 + 1

As per remainder theorem, when f(x) is divided by (x-a), the remainder is f(a).

hence for a = 1

f(a) = 1 + 1 + 1 = 3 (Your answer)

Answer 4

Let (x^31 + x^17 + 1) be represented as f(x).

f(x) = x^31 + x^17 + 1

It's given that f(x) will be divided by (x-1).

x-1 = 0

x = 1

So, x is actually 1.

Sub. (x = 1) into f(x):

f(1) = (1)^31 + (1)^17 + 1

f(1) = 1 + 1 + 1

f(1) = 3

Hence, the remainder is 3.