Q4 of 31 Page 245

In each pair of polynomials below find what kind of natural number n must be, so that the first is a factor of the second.

x2 – 1, xn – 1

Given, x2 – 1, xn – 1 pair of polynomials


Need to find out n such as first polynomial is factor of second


⇒ To check x2 – 1 is factor of xn – 1 we must get xn – 1 = 0 when substituted with x values from the first polynomial


⇒ Here, x2 – 1 so, x = √1 = 1


⇒ Consider n as 1 then the polynomial equation will be x1 – 1 which is equal to zero and x2 – 1 is the factor.


We get, 1 – 1 = 0


∴ x2 – 1 is the factor of xn – 1


Hence, n = 1


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