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Math Help - Help with factoring

  1. #1
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    Help with factoring

    x^4-6x^2-8x-3

    I tried using synthetic division but the graphs of the two were no where close
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  2. #2
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    (x-3) should be a root. Go from there.
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  3. #3
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    Hello, xclo0sive!

    What does synthetic division have to do with graphs?


    Factor: . f(x) \:=\:x^4-6x^2-8x-3
    Did you notice that f(\text{-}1) \:=\:0 ?
    . . This means (x+1) is a factor of f(x).

    \text{We have: }\;f(x) \;=\;(x+1)\underbrace{(x^3 - x^2 - 5x - 3)}_{g(x)}

    Note that g(\text{-}1) \:=\:0
    . . Hence: (x+1) is a factor of g(x).

    \text{We have: }\;f(x) \;=\;(x+1)(x+1)\underbrace{(x^2-2x-3)}_{h(x)}


    Finally, we see that h(x) can be factored.


    Therefore: . f(x) \:=\:(x+1)(x+1)(x+1)(x-3) \:=\:(x+1)^3(x-3)

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  4. #4
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    well i assumed that the factored form should have the same graph as the original equation but i didnt get the same graph. I have one question, what did you do to get (x+1)(x^3 - x^2 - 5x - 3)?
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  5. #5
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    Since you know one root, you can use synthetic division on the polynomial. Also, you know that the only rational solutions are 1, -1, 3, -3, so you can look at those first, and see which of those are roots.
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