The HCF of x^{8}-1 and x^{4}+2x^{3}-2x-1 is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) To find the $HCF$,we factorize both expressions.First expression: $x^{8}-1 = (x^{4}-1)(x^{4}+1) = (x^{2}-1)(x^{2}+1)(x^{4}+1) = (x-1)(x+1)(x^{2}+1

Vedclass · Accepted Answer

$x^{2}-1$

Vedclass · Answer

(C) To find the $HCF$,we factorize both expressions.First expression: $x^{8}-1 = (x^{4}-1)(x^{4}+1) = (x^{2}-1)(x^{2}+1)(x^{4}+1) = (x-1)(x+1)(x^{2}+1)(x^{4}+1)$.Second expression: $x^{4}+2x^{3}-2x-1 = (x^{4}-1) + 2x(x^{2}-1) = (x^{2}-1)(x^{2}+1) + 2x(x^{2}-1) = (x^{2}-1)(x^{2}+2x+1) = (x-1)(x+1)(x+1)^{2} = (x-1)(x+1)^{3}$.The common factors are $(x-1)$ and $(x+1)$.Therefore,the $HCF = (x-1)(x+1) = x^{2}-1$.

The $HCF$ of $x^{8}-1$ and $x^{4}+2x^{3}-2x-1$ is

Similar Questions

Seema,Meena,and Reena start jogging around a circular stadium and complete one round in $54 \text{ seconds}$,$42 \text{ seconds}$,and $63 \text{ seconds}$ respectively. Approximately after how many minutes will they meet again at the starting point?

The $LCM$ of two prime numbers $x$ and $y$ $(x > y)$ is $161$. The value of $3y - x$ is

Find the greatest number of $4$ digits and the least number of $5$ digits that have $196$ as their $HCF$.

The traffic lights at three different road crossings change after $24$ seconds,$36$ seconds,and $54$ seconds,respectively. If they all change simultaneously at $10:15:00$ am,then at what time will they again change simultaneously?

Find the two largest numbers of four digits having $531$ as their $H.C.F$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module