Results 1 to 4 of 4

Thread: polynomial division + simultaneous equations + other :S

  1. #1
    Newbie
    Joined
    Sep 2008
    From
    UK
    Posts
    12

    polynomial division + simultaneous equations + other :S

    Sorry about ridiculous title, but i wasnt sure ....

    1) it is given that $\displaystyle f(x)= xcubed+3xsquared-6x-8$ (having large problems with the maths layout thing)

    hence express $\displaystyle f(x)$ as a product of 3 linear factors

    2)Solve $\displaystyle 4a+2b+12=0 $and$\displaystyle a-b+3=0$

    The 0's are really throwing me on this one :S
    Last edited by vanpopta; Oct 2nd 2008 at 12:54 PM. Reason: maths box is being awkward :S
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    1. $\displaystyle f(x) = x^3 + 3x^2 - 6x - 8$

    Try factoring out $\displaystyle x - 2$.

    $\displaystyle x^2(x - 2) = x^3 - 2x^2$

    $\displaystyle x^3 + 3x^2 - 6x - 8$
    $\displaystyle -(x^3 - 2x^2)$
    $\displaystyle = 5x^2 - 6x - 8$

    $\displaystyle 5x(x - 2) = 5x^2 - 10x$

    $\displaystyle 5x^2 - 6x - 8$
    $\displaystyle -(5x^2 - 10x)$
    $\displaystyle = 4x - 8$

    $\displaystyle 4(x - 2) = 4x - 8$

    $\displaystyle 4x - 8$
    $\displaystyle -(4x - 8)$
    $\displaystyle = 0$

    $\displaystyle \frac{x^3 + 3x^2 - 6x - 8}{x - 2} = x^2 + 5x + 4$

    $\displaystyle x^3 + 3x^2 - 6x - 8 = (x - 2)(x^2 + 5x + 4) = (x - 2)(x + 1)(x + 4)$

    2. Change the system to
    $\displaystyle 4a + 2b = -12$
    $\displaystyle a - b = -3$.

    Multiplying the second equation by 2 yields
    $\displaystyle 4a + 2b = -12$
    $\displaystyle 2a - 2b = -6$

    Adding those two equations yields
    $\displaystyle 6a = -18$.

    Can you finish?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2008
    From
    UK
    Posts
    12
    1st one = wow! so complicated

    2nd one = i feel stupid :P


    Thanks so much
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    The process I used for solving the first problem is known as long division of polynomials (and a little bit of factoring). For polynomials of degree 3 or greater, you will have to try dividing the polynomial by different factors using either this method or synthetic division, which is basically a symbolic representation of long division.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Division by a polynomial
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: Jul 31st 2011, 11:19 PM
  2. Polynomial division vs synthetic division
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Apr 9th 2009, 05:49 AM
  3. Replies: 3
    Last Post: Feb 27th 2009, 07:05 PM
  4. polynomial division
    Posted in the Algebra Forum
    Replies: 6
    Last Post: Nov 13th 2008, 04:02 AM
  5. polynomial division
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Mar 31st 2008, 12:17 AM

Search Tags


/mathhelpforum @mathhelpforum