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Homework Help: Find the real zeros of the function algebraically

  1. Aug 31, 2010 #1
    f(x)=x^3+5x^2+9x+45
    x^3+5x^2+9x+45=0
    x^2(x+5)+9(x-5)=0
    (x^2+9)(x+5)=0
    What happends to the (x+5)?
     
  2. jcsd
  3. Sep 1, 2010 #2

    Mark44

    Staff: Mentor

    There's an error in the step above.
    Since (x2 + 9)(x + 5) = 0, then either x2 + 9 = 0 or x + 5 = 0. Can you continue from here?
     
  4. Sep 1, 2010 #3

    eumyang

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    Homework Helper

    [tex]\begin{aligned}
    x^3 + 5x^2 + 9x + 45 &= 0 \\
    x^2(x + 5) + 9(x + 5) &= 0 \\
    (x^2 + 9)(x + 5) &= 0
    \end{aligned}[/tex]
    I'm assuming that the error pointed out above is just a typo, so I'll just fix it. As for
    It was the greatest common factor, so it got factored out.

    Say you wanted to factor 16x - 28. The GCF is 4, and you could rewrite the expression as
    4(4x) - 4(7),
    and then factor out the 4 to get
    4(4x - 7).

    It's the same thing in your problem. The GCF is an expression: x + 5. You factor the x + 5 out in [tex]x^2 \bold{(x + 5)} + 9 \bold{(x + 5)} = 0[/tex] and you get [tex](x^2 + 9)(x + 5) = 0[/tex]. Maybe part of the confusion is that the GCF appears last instead of first?


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    Last edited: Sep 1, 2010
  5. Sep 1, 2010 #4

    HallsofIvy

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    Science Advisor

    It was the greatest linear factor. It is not a monomial, it is a binomial, just like [itex]x^2+ 9[/itex].

     
  6. Sep 1, 2010 #5

    eumyang

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    Homework Helper

    Fixed. Didn't mean to use the term "monomial" -- just GCF. This is what you get for replying at 5am. :P


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