# Thread: hcf of two polynomials

1. ## hcf of two polynomials

Let $P_1=x^3-3x-2$ and $P_2=x^4+x^3+2x^2+5x+3$. Find $hcf (P_1, P_2).$
Is this a short way of doing the question:

$P_1(x)=x^3-3x-2=(x+1)^2(x-2)$
$P_2(x)=x^4+x^3+2x^2+5x+3=(x+1)^2(x^2-x+3)$

Hence $hcf(P_1,P_2)=(x+1)^2.$

2. Hello,
Originally Posted by Showcase_22
Is this a short way of doing the question:

$P_1(x)=x^3-3x-2=(x+1)^2(x-2)$
$P_2(x)=x^4+x^3+2x^2+5x+3=(x+1)^2(x^2-x+3)$

Hence $hcf(P_1,P_2)=(x+1)^2.$
It is correct to factor this way !

However, note that $k (x+1)^2$, where $k \in (-\infty,0) \cup (0,+\infty)$ also divides the two polynomials.

This is the difference with polynomials : you can't stick to unit (leading coeff = 1) polynomials as hcf.
Any multiple of a common factor is also a common factor.

3. Thanks Moo (lifesaver as always!)

hmm, is it always necessary to include k? It makes sense being there but my lecturer didn't use it.

4. Originally Posted by Showcase_22
Thanks Moo (lifesaver as always!)

hmm, is it always necessary to include k? It makes sense being there but my lecturer didn't use it.
Actually yep (I checked on the wikipedia), because we want a unique hcf, so there are two possibilities :
k=1 or k=gcd(leading coefficients of the polynomials)

Sorry for the confusion :s
But I hope it'll be useful for ya in the future