Question

If the LCM of the polynomials $x^5+x^4+x^2+1$ and $x^4-1$ is $x^8-1$, then their $\mathrm{HCF}$ is

• A. $x^2+1$
• B. $x^4+1$
• C. $x^2-1$
• D. $x^4-1$

Answer 1

Answer: (C)

We have, $P(x)Q(x)=\mathrm{LCM}\times \mathrm{HCF}$
$\left(x^8+x^4+x^2+1\right)\left(x^4-1\right)=\left(x^8-1\right)\times \mathrm{HCF}$
$\therefore \mathrm{HCF}=\frac{\left(x^8+x^4+x^2+1\right)-\left(x^4-1\right)}{\left(x^4-1\right)\left(x^4+1\right)}$
$=\frac{x^8+x^4+x^2+1}{x^4+1}$