Question

which of the following is not a factor of x^17-x : (1) x^2-1 (2) x^6+1 (3) x^4-1 (4) x^8+1​

Answer 1

Answer:

Apply remainder theorem  

x+1=0

x=−1

Put the value of x=−1 in all equations.  

(i) x3+x2+x+1=(−1)3+(−1)2+(−1)+1=−1+1−1+1=0

Then x+1 is the factor of equation

(ii) x4+x3+x2+x+1=(−1)4+(−1)3+(−1)2+(−1)+1=1−1+1−1+1=1

This is not zero.Then x+1 is not the factor of equation

(iii) x4+3x3+3x2+x+1=(−1)4+3(−1)3+3(−1)2+(−1)+1=1

This is not zero.Then x+1 is not the factor of equation

(iv)x3−x2−(2+2)x+

Step-by-step explanation:

Answer 2

Concept

First we will try to calculate the factors of x^17-x by taking x common from ir and then compare it with the given option to check each one of them whether these factors match with our results or not.

Given

The given function is x^17-x.

Find

We have to calculate which factor of the given option is not of x^17-x.

Solution

Since, we have x^17-x

Therefore taking x from from it, we have

x^17-x=x(x^16-1)

=x( (x^8)^2-(1^8)^2)

=x(x^8-1)(x^8+1)

Since we know that (a^2-b^2)=(a-b)(a+b), now

x^17-x=x( (x^4)^2-1)(x^8+1)

=x(x^4-1)(x^4+1)(x^8+1)

=x( (x^2)^2 -1)(x^4+1)(x^8+1)

=x(x^2+1)(x^2-1)(x^4+1)(x^8+1)

Therefore from analysing the above equation we see that x^6+1 didn’t come in any step.

Hence x^6+1 is not the factor of x^17-x. So, option 2 is the correct option.

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