Results 1 to 5 of 5

Math Help - Remainder Theorem

  1. #1
    Member
    Joined
    Oct 2006
    Posts
    185
    Thanks
    1

    Remainder Theorem

    Find the remainder when x^{100}+5 is divided by x^2-2x+1

    Anyone can help? Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823
    Quote Originally Posted by acc100jt View Post
    Find the remainder when x^{100}+5 is divided by x^2-2x+1

    Anyone can help? Thanks
    x^2-2x+1=(x-1)(x-1)

    The remainder theorem states that if a polynomial f(x) is divided by x-k, the remainder is r=f(k).

    So, if I were to divide x^{100}+5 by (x-1), this implies that the remainder is (1)^{100}+5=6. If I were to do it again...

    I think. It's been a long time man.
    Last edited by VonNemo19; July 26th 2009 at 08:11 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2006
    Posts
    185
    Thanks
    1
    i understand the first part. but what do u mean "do it again"?what is the dividend if i wan to divide again?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823
    Quote Originally Posted by acc100jt View Post
    i understand the first part. but what do u mean "do it again"?what is the dividend if i wan to divide again?
    By the Division Algorithm

    f(x)=d(x)q(x)+r(x)

    The way to use this is

    \frac{f(x)-r(x)}{d(x)}=q(x)

    where f(x)=x^{100}+5,r(x)=6,\text{ and },q(x)=\text{ the new function to be divided}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2006
    Posts
    185
    Thanks
    1
    this is what I've tried,

    Let f(x)=x^{100}+5

    f(1)=6

    So
    f(x)=(x-1)Q(x)+6....(1)
    f(x)-6=(x-1)Q(x)
    x^{100}-1=(x-1)Q(x)
    (x-1)(x^{99}+x^{98}+...+x+1)=(x-1)Q(x)

    Thus Q(x)=x^{99}+x^{98}+...+x+1

    Since Q(1)=100
    So Q(x)=(x-1)P(x)+100, substitute this into (1)

    f(x)=(x-1)[(x-1)P(x)+100]+6
    f(x)=(x-1)^{2}P(x)+100x-94

    Hence the remainder is 100x-94

    Am I correct?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. remainder theorem
    Posted in the Algebra Forum
    Replies: 3
    Last Post: October 9th 2010, 01:29 AM
  2. remainder theorem
    Posted in the Algebra Forum
    Replies: 3
    Last Post: September 17th 2009, 05:26 AM
  3. Remainder Theorem
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: July 19th 2009, 02:08 PM
  4. remainder theorem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 9th 2009, 01:16 PM
  5. Factor Theorem and Remainder Theorem
    Posted in the Algebra Forum
    Replies: 2
    Last Post: September 8th 2007, 10:50 AM

Search Tags


/mathhelpforum @mathhelpforum