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Math Help - Factoring

  1. #1
    Junior Member Freaky-Person's Avatar
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    Factoring

    Basically what I need is an explenation and the steps to solve two different types of factoring problems.

    I could care less about the answer, I have it in the back of the sheet anyway.

    1. x^3 - 6x^2 + 11x - 6 = 0


    2. 4(x^3/2)(y^2/3) + 18(x^1/2)(y^5/3)

    Once again I need the steps for the first one, to help factor that kind, and the explenation with the fraction factoring.
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  2. #2
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    Quote Originally Posted by Freaky-Person View Post
    1. x^3 - 6x^2 + 11x - 6 = 0
    \begin{aligned}x^3-6x^2+11x-6&=x^3-(x^2+5x^2)+(5x+6x)-6\\<br />
&=(x^3-x^2)-(5x^2-5x)+(6x-6)\\<br />
&=x^2(x-1)-5x(x-1)+6(x-1)\\<br />
&=(x-1)(x^2-5x+6)\\<br />
&={\color{blue}(x-1)(x-2)(x-3)}\,\blacksquare<br />
\end{aligned}
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  3. #3
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    Quote Originally Posted by Freaky-Person View Post
    2. 4(x^3/2)(y^2/3) + 18(x^1/2)(y^5/3)
    4x^{3/2}y^{2/3}+18x^{1/2}y^{5/3}=4x^{1/2}\cdot x\cdot y^{2/3}+18x^{1/2}\cdot y^{2/3}\cdot y

    Does that make sense?
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  4. #4
    Junior Member Freaky-Person's Avatar
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    Quote Originally Posted by Krizalid View Post
    4x^{3/2}y^{2/3}+18x^{1/2}y^{5/3}=4x^{1/2}\cdot x\cdot y^{2/3}+18x^{1/2}\cdot y^{2/3}\cdot y

    Does that make sense?
    Alrighty.. No I didn't get that.

    All you really needed to do was factor out the correct amount... I just didn't get what amount that would be with the fractions..

    If it helps, the answer is -> [2(x^1/2)(y^2/3)][2x+9y]
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  5. #5
    Junior Member Freaky-Person's Avatar
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    Quote Originally Posted by Krizalid View Post
    \begin{aligned}x^3-6x^2+11x-6&=x^3-(x^2+5x^2)+(5x+6x)-6\\<br />
&=(x^3-x^2)-(5x^2-5x)+(6x-6)\\<br />
&=x^2(x-1)-5x(x-1)+6(x-1)\\<br />
&=(x-1)(x^2-5x+6)\\<br />
&={\color{blue}(x-1)(x-2)(x-3)}\,\blacksquare<br />
\end{aligned}
    That great and all, but I don't get WHY you seperated the x^2 and x values, and why by that amount.
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by Freaky-Person View Post
    Basically what I need is an explenation and the steps to solve two different types of factoring problems.

    I could care less about the answer, I have it in the back of the sheet anyway.

    1. x^3 - 6x^2 + 11x - 6 = 0
    The rational root theorem tells us that any rational root of a polynomial is a
    +/- a factor of the constant term divided by +/- a factor of the coeficient of the highest power.

    So in this case the rational root theorem tells us that any rational roots
    of this cubic are amoung 1, -1, 2, -2, 3, -3.

    Try x=3 in x^3 - 6x^2 + 11x - 6 gives 0, so x-3 is a factor of x^3 - 6x^2 + 11x - 6.

    Now dividing x^3 - 6x^2 + 11x - 6 by x-3 gives:

    <br />
x^3 - 6x^2 + 11x - 6=(x-3)(x^2-3x+2)<br />

    and the last term can be factored by inspection, or finding the roots using
    the quadratic formula or the same procedure as before to get:

    <br />
x^3 - 6x^2 + 11x - 6=(x-3)(x^2-3x+2)=(x-3)(x-1)(x-2)<br />

    RonL
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by Freaky-Person View Post
    2. 4(x^3/2)(y^2/3) + 18(x^1/2)(y^5/3)
    Take out the common factor 2 x^{1/2}y^{2/3} to get:

    <br />
4x^{3/2}y^{2/3} + 18x^{1/2}y^{5/3}= 2 x^{1/2}y^{2/3}(2 x +9y)<br />

    RonL
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