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Math Help - Factor and Solve

  1. #1
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    Post Factor and Solve

    My first question is f(x)=9x^4-27
    *must factor out completly (don't forget to include complex roots)

    ---- i don'yt know where to start. i can't use the quadratic---since it is raised to the 4th how would i factor out? would it be
    9x^4+27x^2-27-27 but then i have a x^2 in my equation.

    my second question is
    Solve 3x^2+7x=-45

    ---- if i use the quatric in this equation i get a decimal under my root. So i'm stuck after i equal it to zero.
    3x^2+7x+45=0
    i'm having trouble factoring this out. i need 2 numbers that when mult equal 45 and when added/subtracted equal 7? but both of my signs need to be (+) positive?

    Help please
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  2. #2
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    Quote Originally Posted by b_austintx View Post
    My first question is f(x)=9x^4-27
    *must factor out completly (don't forget to include complex roots)

    ---- i don'yt know where to start. i can't use the quadratic---since it is raised to the 4th how would i factor out? would it be
    9x^4+27x^2-27-27 but then i have a x^2 in my equation.

    my second question is
    Solve 3x^2+7x=-45

    ---- if i use the quatric in this equation i get a decimal under my root. So i'm stuck after i equal it to zero.
    3x^2+7x+45=0
    i'm having trouble factoring this out. i need 2 numbers that when mult equal 45 and when added/subtracted equal 7? but both of my signs need to be (+) positive?

    Help please

    The first one is repeated use of the Difference of Two Squares.

    9x^4 - 27 = 9(x^4 - 3)

     = 9\left[(x^2)^2 - \left(3^{\frac{1}{2}}\right)^2\right]

     = 9\left(x^2 - 3^{\frac{1}{2}}\right)\left(x^2 + 3^{\frac{1}{2}}\right)

     = 9\left[x^2 - \left(3^{\frac{1}{4}}\right)^2\right]\left[x^2 + \left(3^{\frac{1}{4}}\right)^2\right]

     = 9\left[x + 3^{\frac{1}{4}}\right]\left[x - 3^{\frac{1}{4}}\right]\left[x^2 - \left(3^{\frac{1}{4}}i\right)^2\right]

     = 9\left[x + 3^{\frac{1}{4}}\right]\left[x - 3^{\frac{1}{4}}\right]\left[x + 3^{\frac{1}{4}}i\right]\left[x - 3^{\frac{1}{4}}i\right].
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  3. #3
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    Quote Originally Posted by b_austintx View Post
    My first question is f(x)=9x^4-27
    *must factor out completly (don't forget to include complex roots)

    ---- i don'yt know where to start. i can't use the quadratic---since it is raised to the 4th how would i factor out? would it be
    9x^4+27x^2-27-27 but then i have a x^2 in my equation.
    I would use " a^2- b^2= (a+ b)(a- b)". Here a= 3x, obviously. 27 doesn't look like "b^2" but 27= (\sqrt{27})^2= (3\sqrt{3})^2.

    9x^4- 27= (3x^2- 3\sqrt{3})(3x^2+ 3\sqrt{3}).

    We can do the same thing with 3x^2- 3\sqrt{3}: it is of the form a^2- b^2 with a= x\sqrt{3} and b= \sqrt{3}\sqrt[4]{3}. 3x^2- 3\sqrt{3}= (x\sqrt{3}+ \sqrt{3}\sqrt[4]{3})(x\sqrt{3}- \sqrt{3}\sqrt[4]{3})= 3(x+ \sqrt[4]{3})(x- \sqrt[4]{3}).

    3x^2+ 3\sqrt{3}= 3x^2- (-1)(3)\sqrt{3} only requires that we put in [itex]-1= i^2[/itex]. [tex]3x^2+ 3\sqrt{3}= 3(x+ i\sqrt[4]{3})(x- i\sqrt[4]{3}).

    Putting those together, 9x^4- 27= 9(x+ \sqrt[4]{3})(x- \sqrt[4]{3})(x+ i\sqrt[4]{3})(x- i\sqrt[4]{3})

    Having done that, it occurs to me that the easiest way to factor a polynomial is to set it equal to 0 and solve the equation! If 9x^4- 27=  9(x^4- 3)= 0, then x^4= 3 has roots \sqrt[4]{3}, -\sqrt[4]{3}, i\sqrt[4]{3}, and -i\sqrt[4]{3} which tells us that x^4- 3= (x- \sqrt[4]{3})(x+ \sqrt[4]{3})(x- i\sqrt[4]{3})(x+ i\sqrt[4]{3}) and so 9x^4- 27= 9(x- \sqrt[4]{3})(x+ \sqrt[4]{3})(x- i\sqrt[4]{3})(x+ i\sqrt[4]{3}) as above.
    my second question is
    Solve 3x^2+7x=-45

    ---- if i use the quatric in this equation i get a decimal under my root. So i'm stuck after i equal it to zero.
    3x^2+7x+45=0
    i'm having trouble factoring this out. i need 2 numbers that when mult equal 45 and when added/subtracted equal 7? but both of my signs need to be (+) positive?

    Help please
    quartic? This is quadratic and the quadratic formula works nicely: 3x^2+ 7x+ 45= 0 has roots x= \frac{-7\pm\sqrt{49-540}}{6}= \frac{-7\pm\sqrt{-491}}{6}= \frac{-7\pm i\sqrt{491}}{6}. 491 happens to be prime so that square root cannot be reduced but we can write 3x^3+ 7x+ 45= 3(x+\frac{7}{6}- i\frac{\sqrt{491}}{6})(x+\frac{7}{6}+ i\frac{\sqrt{491}}{6})
    Last edited by HallsofIvy; April 28th 2009 at 05:45 PM.
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