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Math Help - Factoring Questions <3 pplz help exam tomorrow :(

  1. #1
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    Factoring Questions <3 pplz help exam tomorrow :(

    i've been cramming all day... im so tired! i still have to learn so much more

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  2. #2
    MHF Contributor red_dog's Avatar
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    1) 20x^9y^9-8x^6y^7+36x^3y^5=4x^3y^5(5x^6y^4-2x^3y^2+9)

    2), 3), 4):
    aX^2+bX+c=a(X-x_1)(X-x_2)
    where x_1, \ x_2 are the roots of the quadratic ax^2+bx+c=0.

    I'll do it for 2):
    15z^2+14z-8=0\Rightarrow z_1=-\frac{4}{3}, \ z_2=\frac{2}{5}
    Then 15z^2+14z-8=15\left(z+\frac{4}{3}\right)\left(z-\frac{2}{5}\right)=\left[3\left(z+\frac{4}{3}\right)\right]\left[5\left(z-\frac{2}{5}\right)\right]=(3z+4)(5z-2)

    Now you can do the others.
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  3. #3
    Super Member angel.white's Avatar
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    Quote Originally Posted by red_dog View Post
    1) 20x^9y^9-8x^6y^7+36x^3y^5=4x^3y^5(5x^6y^4-2x^3y^2+9)

    2), 3), 4):
    aX^2+bX+c=a(X-x_1)(X-x_2)
    where x_1, \ x_2 are the roots of the quadratic ax^2+bx+c=0.

    I'll do it for 2):
    15z^2+14z-8=0\Rightarrow z_1=-\frac{4}{3}, \ z_2=\frac{2}{5}
    Then 15z^2+14z-8=15\left(z+\frac{4}{3}\right)\left(z-\frac{2}{5}\right)=\left[3\left(z+\frac{4}{3}\right)\right]\left[5\left(z-\frac{2}{5}\right)\right]=(3z+4)(5z-2)

    Now you can do the others.
    Wow, I never knew that, that is neat!

    (always seemed like there ought to be a way to use the roots in factoring, but I never actually knew how)

    Also, your avatar looks like a firebat from starcraft O.o
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  4. #4
    Math Engineering Student
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    Another attempt is to note that 20-6=14

    So

    \begin{aligned}<br />
15z^2 + 14z - 8 &= 15z^2 + (20z - 6z) - 8\\<br />
&= \left( {15z^2 + 20z} \right) - (6z + 8)\\<br />
&= 5z(3z + 4) - 2(3z + 4)\\<br />
&= (3z + 4)(5z - 2).<br />
\end{aligned}
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