Results 1 to 2 of 2

Math Help - Dividing equations and finding quotiant.

  1. #1
    Newbie
    Joined
    Apr 2010
    Posts
    15

    Dividing equations and finding quotiant.

    i)Find the quotient and the remainder when 3x^3-2x^2+x+7 is divided by x^2-2x+5#

    ii)Hence, or otherwise, determine the values of the constants a and b such that, 3x^3-2x^2+ax+b is divded by x^2-2x+5, there is no remainder.


    Been looking at this for ages and have forgotten how to dot his kind of quiestion.
    Please help!
    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,307
    Thanks
    1284
    Quote Originally Posted by George321 View Post
    i)Find the quotient and the remainder when 3x^3-2x^2+x+7 is divided by x^2-2x+5#

    ii)Hence, or otherwise, determine the values of the constants a and b such that, 3x^3-2x^2+ax+b is divded by x^2-2x+5, there is no remainder.


    Been looking at this for ages and have forgotten how to dot his kind of quiestion.
    Please help!
    Thanks
    Essentially, the same as long division of numbers:
    Look at the leading term. x^2 divides into 3x^3 3x times. Now multiply the entire divisor, x^2- 2x+ 5 by that and subtract:
    \begin{array}{cccc}3x^3& - 2x^2& + x& + 7 \\ \underline{3x^3}& \underline{- 6x^2}& \underline{+ 15x} &  \\ & 4x^2 & -14x & 7\end{array}

    Now, x^2 will divide into 4x^2 4 times. Multiply x^2- 2x+ 5 by 4 and subtract:
    \begin{array}{ccc}4x^2 & -14x & 7 \\\underline{4x^2}& \underline{-8x} & \underline{20}\\ & -6x& - 13\end{array}.

    That is, x^2- 2x+ 5 divides into 3x^3- 2x^2+ x+ 7 3x+ 4 times with remainder -6x- 13.

    For the second problem, do the same thing but with "ax+ b" rather than "x+ 7". The remainder will involve a and b- set the coefficient of x and the constant term equal to 0 and solve for a and b.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. finding equations
    Posted in the Algebra Forum
    Replies: 4
    Last Post: March 8th 2010, 02:42 PM
  2. Finding equations
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 4th 2009, 11:39 PM
  3. Rational Equations- Multiplying and dividing
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 2nd 2009, 10:25 PM
  4. Replies: 2
    Last Post: August 12th 2008, 06:54 PM
  5. Replies: 1
    Last Post: September 1st 2007, 06:35 AM

Search Tags


/mathhelpforum @mathhelpforum