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Math Help - Factoring To Find Absolute Max/Min?

  1. #1
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    Exclamation Factoring To Find Absolute Max/Min?

    How do I go about factoring this correctly? (the bold)


    f(x)= 2x^3-x^2-4x+8

    F^1(x) = 3x^2-2x-4

    (3x ) (x ) ?
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  2. #2
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    Quote Originally Posted by JumpinJimmy View Post
    How do I go about factoring this correctly? (the bold)


    f(x)= 2x^3-x^2-4x+8

    F^1(x) = 3x^2-2x-4

    (3x ) (x ) ?
    Dear JumpinJimmy,

    Do you mean that you have to factorize, f(x)=3x^{2}-2x-4???
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    yeah sorry about the confusion, do not know the short cuts for subscripts and superscripts.
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    Quote Originally Posted by JumpinJimmy View Post
    yeah sorry about the confusion, do not know the short cuts for subscripts and superscripts.
    Dear JumpinJimmy,

    3x^{2}-2x-4

    =3\left(x^{2}-\frac{2}{3}x-\frac{4}{3}\right)

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

    =3\left(\left(x-\frac{1}{3}\right)^{2}-\frac{13}{9}\right)

    =3\left(\left(x-\frac{1}{3}\right)^{2}-\left(\frac{\sqrt{13}}{3}\right)^{2}\right)

    =3\left(x-\frac{1}{3}-\frac{\sqrt{13}}{3}\right)\left(x-\frac{1}{3}+\frac{\sqrt{13}}{3}\right)

    =3\left(x-\frac{(1-\sqrt{13})}{3}\right)\left(x-\frac{(1+\sqrt{13})}{3}\right)

    Hope this will help you.
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  5. #5
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    Quote Originally Posted by JumpinJimmy View Post
    How do I go about factoring this correctly? (the bold)


    f(x)= 2x^3-x^2-4x+8

    F^1(x) = 3x^2-2x-4

    (3x ) (x ) ?
    The derivative is actually

    f'(x) = 6x^2 - 2x - 4.


    So if you are setting it equal to 0 and factorising

    6x^2 - 2x - 4 = 0

    2(3x^2 - x - 2) = 0

    2(3x^2 - 3x + 2x -2) = 0

    2[3x(x - 1) + 2(x - 1)] = 0

    2(x - 1)(3x + 2) = 0


    So x - 1 = 0 or 3x + 2 = 0

    x = 1 or x = -\frac{2}{3}.
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