Highest common factor Watch

Substitution
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ind the quotient and remainder when f(X)=X^3+4⋅X^2+3⋅X+15 is divided by g(X)=X^2+3. Hence, find hcf(f(X),g(X)).

I calculated the quorient to be x+4 and the remainder 3/(x^2+3) but not sure how to calculate the hcf.

Thanks!
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Zacken
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(Original post by Substitution)
ind the quotient and remainder when f(X)=X^3+4⋅X^2+3⋅X+15 is divided by g(X)=X^2+3. Hence, find hcf(f(X),g(X)).

I calculated the quorient to be x+4 and the remainder 3/(x^2+3) but not sure how to calculate the hcf.

Thanks!
Would it not simply be 1?
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Substitution
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(Original post by Zacken)
Would it not simply be 1?
Maybe, is that because a remainder exists when dividing?
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Zacken
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(Original post by Substitution)
Maybe, is that because a remainder exists when dividing?
I think it's because x^2 + 3 is irreducible and the reason you've given.

Edit: I'm going to add in a disclaimer that I don't really know what I'm talking about here - so somebody fele free to jump in and correct me. :-)
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poorform
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(Original post by Substitution)
ind the quotient and remainder when f(X)=X^3+4⋅X^2+3⋅X+15 is divided by g(X)=X^2+3. Hence, find hcf(f(X),g(X)).

I calculated the quorient to be x+4 and the remainder 3/(x^2+3) but not sure how to calculate the hcf.

Thanks!
So you have \displaystyle X^3+4X^2+3X+15=(X+4)(X^2+3)+3.

Now use the following result let \displaystyle f(X),g(X),q(X),r(X) \in \mathbb{F}[X].

If \displaystyle f(X)=q(X)g(X)+r(X)

then \displaystyle \text{hcf}(f(X),g(X))=\text{hcf}  (g(X),r(X))

Now from above we have \displaystyle \text{hcf}(X^3+4X^2+3X+15,X^2+3)  =\text{hcf}(X^2+3,3)=1.

In fact for \displaystyle f(X) \in \mathbb{F}[X], p \in \mathbb{F}, p \neq 0, ~\text{hcf}(f(X),p)=1
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