Results 1 to 4 of 4

Math Help - hcf of two polynomials

  1. #1
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,
    Quote Originally Posted by Showcase_22 View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by Showcase_22 View Post
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 7
    Last Post: April 7th 2011, 12:38 PM
  2. GCD of polynomials in Zn[x]
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: May 18th 2010, 06:22 AM
  3. Polynomials
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 16th 2010, 06:52 AM
  4. Replies: 7
    Last Post: January 8th 2010, 03:13 AM
  5. Replies: 5
    Last Post: November 29th 2005, 03:22 PM

Search Tags


/mathhelpforum @mathhelpforum