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Factoring a polynomial

  1. Sep 10, 2009 #1
    1. The problem statement, all variables and given/known data

    8x^3 + 4X^2 - 2x -1

    2. Relevant equations

    remainder theorem
    factor theorem


    3. The attempt at a solution

    None yet, not sure how to go about it when there is a coefficent in front of the highest power of X.
     
  2. jcsd
  3. Sep 10, 2009 #2

    rock.freak667

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    so if 8x3 + 4x2 - 2x -1=0 and we divide by 8, the last term is -1/8. What product of fractions will give you -1/8?

    -1/8 and 1 would give -1/8 do either of these satisfy the equation?

    What are the other options to choose from?
     
  4. Sep 10, 2009 #3
    is that the way we do it? Divide all by 8. I thought about that but thought it was wrong.

    Just noticed you made them equal to zero. Why did you do that? The original question does not make them equal to zero.

    I divided all by 8 and noticed neither -1/8 nor 1 satisfy the remainder theorem

    Also, i cant seem to find a number that makes them zero when using the remainder theorem. I need some more help, sorry.
     
    Last edited: Sep 10, 2009
  5. Sep 10, 2009 #4

    rock.freak667

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    Well we don't need to divide by 8, but we are looking at solution of the form ax+b for when the coefficient of the highest degree in the polynomial is not 1.

    to make -1/8...you can multiply -1/2 and 1/4 or -1/4 and 1/2. Do any of these make the equation zero?
     
  6. Sep 10, 2009 #5
    f(1/2) works

    So one factor is (x - 1/2)

    I will divide to find the second part:
    Second part = 8x^2 +8x +2

    And i forgot how to factor that, lol. using quadratic formula cause i cant do it in my head right now.
     
  7. Sep 10, 2009 #6
    Here is what i got:

    (x- 1/2) (4x+2) (2x +1) I dont have an answer key is this correct?

    Sorry for asking so much of this forum, i promise one day if i learn i will repay by helping others.
     
  8. Sep 10, 2009 #7

    rock.freak667

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    yes but that can be further simplified as 4x+2 =2(x+1), that 2 can be expanded into the x-1/2 to give integer coefficients.
     
  9. Sep 10, 2009 #8

    Dick

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    That's right. You could also write that as (2x-1)*(2x+1)^2. Factor a 2 out of the second factor and move it into the first. Looking forward to your helping help!
     
  10. Sep 10, 2009 #9

    lurflurf

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    ley y=2x
    8x^3 + 4X^2 - 2x -1
    (2x)^3+(2x)^2-(2x)-1
    y^3+y^2-y-1
    factor by grouping
    gcd(y^3,y^2)=y^2
    gcd(-y,-1)=-1
    (y^2)(y+1)+(-1)(y+1)
    and so on
     
  11. Sep 10, 2009 #10
    Thanks guys i got it now. :)
     
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