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Let $f(x)=x^3+x^2+x+1$ and $g(x)=x^3+1$. Then in $\mathbb{Q}[x]$

  1. $\gcd (f(x),g(x))=x+1$

  2. $ \gcd(f(x),g(x))=x^3-1$

  3. $\operatorname{lcm}(f(x),g(x))=x^5+x^3+x^2+1$

  4. $\operatorname{lcm}(f(x),g(x))=x^5+x^4+x^3+x^2+1$

I know how to find the greatest common divisor(gcd) and the least common multiple(lcm) of numbers. But how can I find the gcd and lcm of polynomials?

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In pretty much the same way -- by using Euclidean algorithm!

Dividing $f(x)$ by g(x) you get quotient 1 and remainder $x^2+x$, so you continue like this: $x^3+x^2+x+1=(x^3+1)\cdot1 + x^2+x$,

$x^3+1=(x^2+x)\cdot x + (-x^2+1)$,

$x^2+x = (-x^2+1)\cdot (-1) + x+1$,

$-x^2+1 = (x+1)\cdot (-x+1)$

so their gcd is $x+1$. Finding lcm now should not be hard.

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  • $\begingroup$ lcm, for example, can be found as $\frac{f(x)g(x)}{gcd(f, g)}$ $\endgroup$ – finalni Oct 16 '14 at 10:15
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$f(x)=x^3+x^2+x+1=(x+1)(x^2+1)$ and $g(x)=x^3+1=(x+1)(x^2-x+1)$ $$lcm(f,g)=(x+1)(x^2+1)(x^2-x+1)$$ $$gcd(f,g)=(x+1)$$

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  • $\begingroup$ There is no option like $lcm(f,g)=(x+1)$ $\endgroup$ – S. Pitchai Murugan Oct 16 '14 at 9:26

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