Results 1 to 4 of 4

Thread: Long Division

  1. #1
    Banned
    Joined
    Dec 2009
    From
    NYC
    Posts
    84

    Long Division

    See attachment for question.

    I got the answer: (3x^3/2) - (5x^2/8) - (7x/16) - (53/32) as the quotient.

    I tried using synthetic division but the divisor is not in the form x - c. So, I could not apply synthetic in this case.

    The remainder I got is -159/32.

    Is any of this correct?
    Attached Thumbnails Attached Thumbnails Long Division-long-division.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member SirOJ's Avatar
    Joined
    Dec 2009
    Posts
    45
    I've done this problem a few times now to make sure and got a quotient of

    (3/2)x^3 - (13/4)x^2 + (35/8)x

    and a remainder of

    -121/8


    Hope that helps
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    849
    Hello, sologuitar!


    Sorry, you messed up somewhere . . .


    . . $\displaystyle \begin{array}{cccccccccccc}
    &&&& \frac{3}{2}x^3 & - & \frac{13}{4}x^2 & + & \frac{35}{8}x & - & \frac{121}{16} \\
    & & -- & -- & -- & -- & -- & -- & -- & -- & -- \\
    2x+3 & ) & 3x^4 &-& 3x^3 &-& x^2 &-& 2x &+& 0 \\
    & & 3x^4 &+& \frac{9}{2}x^3 \\
    & & -- & -- & -- \\
    &&&& -\frac{13}{2}x^3 &-& x^2 \\
    &&&& -\frac{13}{2}x^3 &-& \frac{39}{4}x^2 \\
    &&&& -- & -- & -- \end{array}$
    . . . . . . . . . . . . . . . . . . . . . . . . . . . .$\displaystyle \begin{array}{cccccccccccc}
    &&&&&&\frac{35}{4}x^2 &-& 2x \\
    &&&&&& \frac{35}{4}x^2 &+& \frac{105}{8}x \\
    &&&&&& -- & -- & -- \\
    &&&&&&&&-\frac{121}{8}x &+& 0 \\
    &&&&&&&& -\frac{121}{8}x &-& \frac{363}{16} \\
    &&&&&&&& -- & -- & -- \\
    &&&&&&&&&& \frac{363}{16}
    \end{array}$

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Dec 2009
    From
    NYC
    Posts
    84

    soroban...

    Quote Originally Posted by Soroban View Post
    Hello, sologuitar!


    Sorry, you messed up somewhere . . .


    . . $\displaystyle \begin{array}{cccccccccccc}$$\displaystyle
    &&&& \frac{3}{2}x^3 & - & \frac{13}{4}x^2 & + & \frac{35}{8}x & - & \frac{121}{16} \\
    & & -- & -- & -- & -- & -- & -- & -- & -- & -- \\
    2x+3 & ) & 3x^4 &-& 3x^3 &-& x^2 &-& 2x &+& 0 \\
    & & 3x^4 &+& \frac{9}{2}x^3 \\
    & & -- & -- & -- \\
    &&&& -\frac{13}{2}x^3 &-& x^2 \\
    &&&& -\frac{13}{2}x^3 &-& \frac{39}{4}x^2 \\
    &&&& -- & -- & -- \end{array}$
    . . . . . . . . . . . . . . . . . . . . . . . . . . . .$\displaystyle \begin{array}{cccccccccccc}$$\displaystyle
    &&&&&&\frac{35}{4}x^2 &-& 2x \\
    &&&&&& \frac{35}{4}x^2 &+& \frac{105}{8}x \\
    &&&&&& -- & -- & -- \\
    &&&&&&&&-\frac{121}{8}x &+& 0 \\
    &&&&&&&& -\frac{121}{8}x &-& \frac{363}{16} \\
    &&&&&&&& -- & -- & -- \\
    &&&&&&&&&& \frac{363}{16}
    \end{array}$
    Thank you for pointing out that an error was made. This was a tough problem because of all the fractions in the quotient. Typically I solve long division problems where the divisor is in the form (x - c) meaning the coefficient of x is not > 1.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. long division
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Mar 27th 2011, 12:31 PM
  2. Long division
    Posted in the LaTeX Help Forum
    Replies: 4
    Last Post: Jun 27th 2010, 08:55 PM
  3. long division
    Posted in the Pre-Calculus Forum
    Replies: 8
    Last Post: May 19th 2009, 06:52 AM
  4. Long Division
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Jan 31st 2008, 01:56 PM
  5. Long Division
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Jan 20th 2007, 05:39 AM

Search Tags


/mathhelpforum @mathhelpforum