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Math Help - Factor

  1. #1
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    Factor

    3x^2+5x-7


    I want to make sure this is an error in the book.
    You can't factor this right?
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  2. #2
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    Re: Factor

    Quote Originally Posted by habibixox View Post
    3x^2+5x-7


    I want to make sure this is an error in the book.
    You can't factor this right?
    b^2 - 4ac = 25 - 4(3)(-7) = 109 ... not a perfect square, therefore will not factor over the rationals.
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  3. #3
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    Re: Factor

    thanks
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  4. #4
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    Re: Factor

    Quote Originally Posted by habibixox View Post
    3x^2+5x-7
    I want to make sure this is an error in the book.
    You can't factor this right?
    I can factor it,
    \left( {x - \left[ {\frac{{ - 5}}{6} + \frac{{\sqrt {109} }}{6}} \right]} \right)\left( {x - \left[ {\frac{{ - 5}}{6} - \frac{{\sqrt {109} }}{6}} \right]} \right)
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  5. #5
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    Re: Factor

    Quote Originally Posted by Plato View Post
    I can factor it,
    \left( {x - \left[ {\frac{{ - 5}}{6} + \frac{{\sqrt {109} }}{6}} \right]} \right)\left( {x - \left[ {\frac{{ - 5}}{6} - \frac{{\sqrt {109} }}{6}} \right]} \right)
    I certainly hope you mean \displaystyle 3\left[x - \left(-\frac{5}{6} + \frac{\sqrt{109}}{6}\right)\right]\left[x - \left(-\frac{5}{6} - \frac{\sqrt{109}}{6}\right)\right]...


    \displaystyle \begin{align*} 3x^2 + 5x - 7 &= 3 \left( x^2 + \frac{5}{3}x - \frac{7}{3} \right) \\ &= 3 \left[ x^2 + \frac{5}{3}x + \left( \frac{5}{6} \right)^2 - \left(\frac{5}{6}\right)^2 - \frac{7}{3}\right] \\ &= 3\left[\left(x + \frac{5}{6}\right)^2 - \frac{25}{36} - \frac{84}{36}\right] \\ &= 3\left[\left(x + \frac{5}{6}\right)^2 - \frac{109}{36}\right] \\ &= 3\left[\left(x + \frac{5}{6}\right)^2 - \left(\frac{\sqrt{109}}{6}\right)^2\right] \\ &= 3\left(x + \frac{5}{6} - \frac{\sqrt{109}}{6}\right)\left(x + \frac{5}{6} + \frac{\sqrt{109}}{6}\right) \end{align*}
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