Results 1 to 6 of 6

Math Help - Synthetic division

  1. #1
    Member
    Joined
    Jul 2011
    Posts
    196

    Synthetic division

    Find the values of real numbers a and b such that:

    x^4 +4x^3+ax^2-b is divisible by both (x-1) and (x+2)

    Do I need to extract the factors (x-1) and (x+2) FIRST and then try to solve for a and b or how do I start?

    Also, the answers provided are a = 7 and b = 12 but when I try to divide (x-1) into x^4+4x^3+7x^2-12 I get a remainder of 1 which means that it does not divide it and a and b are incorrect values right?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: Synthetic division

    If you make two horner scheme's, one with divisor 1 and the other one with divisor -2 (and at the end let the remainder =0) then you'll come to two simultaneous equations with variables a and b which you can solve directly.

    (Notice: the coefficient of x is 0).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jul 2011
    Posts
    196

    Re: Synthetic division

    Ahh turns out I was just making errors in my synthetic division :S

    Another method I used to get a = 7 and b = 12 is to just sub 1 and -2 into the original equation for x and solve these two equations simultaneously. Then I subbed these values in and divided by (x-1) and then (x+2) to get zero remainder solutions.

    Why does subbing in these values (1 and -2) for x allow you to find a and b? I can see that it works and I can see that they are derived from the (x-1) and (x+2) terms but I don't see the logic behind it.

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: Synthetic division

    So have you tried it with the horner scheme?
    1) divisor 1, the remainder becomes: -b+a+5
    And this remainder has to be equal to 0 if the polynomial is divisible by the factor (x-1).
    2) divisor -2, the remainder becomes: -b+4a-16
    And this remainder has to be equal to 0 if the polynomial is divisible by the factor (x+2).

    Therefore we get a system of simultaneous equations, if you multiply the second (or first) equation with a factor -1, then you can cancel out the variable b out and so find a and afterwards find b.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jul 2011
    Posts
    196

    Re: Synthetic division

    I am watching a video on it now I had never heard of it before.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: Synthetic division

    But offcourse like you already said a more easy way is just entering the zero's into the equation and so you get the same system:
    1) 1 is a zero, so: 1+4+a-b=0 \Leftrightarrow 5+a-b=0
    2) -2 is a zero, so: 16-32+4a-b=0 \Leftrightarrow -16+4a-b=0

    So the same system .
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Synthetic Division
    Posted in the Algebra Forum
    Replies: 2
    Last Post: July 5th 2010, 06:32 PM
  2. synthetic division
    Posted in the Algebra Forum
    Replies: 5
    Last Post: October 28th 2009, 08:10 AM
  3. Synthetic Division
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: September 2nd 2009, 03:57 PM
  4. Polynomial division vs synthetic division
    Posted in the Algebra Forum
    Replies: 2
    Last Post: April 9th 2009, 06:49 AM
  5. Synthetic Division
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: January 20th 2007, 06:37 AM

Search Tags


/mathhelpforum @mathhelpforum