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Factoring Problem

  1. Feb 18, 2006 #1
    I'm in calculus, but we're having this test on some curve sketing soon and I'm doing some practice problems that the teacher gave us. We're not allowed to use a calculator in this calculus class (except for very rare instances) so we have to know how to sketch graphs with no problem!

    I'm wondering I have factored this correctly. If not, I'm wondering if someone can point me in the right direction.

    Orginal Equation:
    x^3 - 3x^2-6x+8

    My Factors
    (x+2)(x^2-2x+4) - 3x(x+2)

    Turns into ...

    (1-3x)(x+2)(x^2 -2x + 4)

    If anyone could help me out I would appreciate it.

  2. jcsd
  3. Feb 18, 2006 #2


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    Science Advisor

    You can always multiply them back out to see if it gets you where you started.
  4. Feb 18, 2006 #3


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    Staff Emeritus
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    Your original polynomial has degree 3 yet you have 4 factors involving x!

    Hint: [tex](x+2)(x^2-2x+4) - 3x(x+2) [/tex] is correct but
    ab- ca= a(b- c) NOT ab(1-c).
  5. Feb 19, 2006 #4
    Ok, thanks for the help thus far. I was able to remember (through the help of another math friend) that if all the coefficients of the polynomial add up to zero, it is divisble by x-1 and x-1 is one of the factors. Therefore I had to take x-1 and divide it into [tex]x^3 - 3x^2 -6x+8[/tex] and I was given [tex]x^2-2x-8[/tex]. From that I could get the other two factors [tex](x-4)(x+2)[/tex]

    Thanks for looking at it though.
  6. Feb 19, 2006 #5


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    Your factored expression is a 4th degree polynomial while your original expression is a 3rd degree polynomial.

    As HallsofIvy said, [tex](x+2)(x^2-2x+4) - 3x(x+2) [/tex] is correct, and the factored form should be

    (x+2)(2nd degree poly).
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