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Math Help - FACTORING! plz help <3 EXAM TOMORROW!

  1. #1
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    FACTORING! plz help <3 EXAM TOMORROW!

    please help me do these questions and explain <3

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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Noelle View Post
    please help me do these questions and explain <3

    1. 20x^9y^9 - 8x^6y^7 + 36x^3y^5

    from here, the lowest power of x is 3, while of y is 5.. also, the coefficients have 4 as their highest common factor..

    so you would have 20x^9y^9 - 8x^6y^7 + 36x^3y^5 = 4x^3y^5 \left( 5x^6y^4 - 2x^3y^2 + 9 \right)

    if you set u = x^3y^2, then you will have u^2 = (x^3y^2)^2 = x^6y^4, so that the one in the parenthesis would become 5u^2 - 2u + 9, which is not factorable..
    hence,

    20x^9y^9 - 8x^6y^7 + 36x^3y^5 = 4x^3y^5 \left( 5x^6y^4 - 2x^3y^2 + 9 \right)

    2. 15z^2 + 14z - 8

    notice that the factors of 15 are 1, 3, 5, 15 (positive and negative) and the factors of 8 are 1,2,4,8 (positive and negative)..

    so what combinations of the factors of 15 and 8 that would give 14?? it is (5)(4) + (3)(-2) = 20 - 6..

    (use the what so called FOIL method..)

    thus you can factor 15z^2 + 14z - 8 to \left( 5z - 2 \right) \left( 3z + 4 \right) ..

    3. do same as number two..

    4. \frac{k^2 +11k + 24}{k^2 + 13k + 40} \cdot \frac{k^2 + 5k}{k^2 - 3k - 18}

    so if you do FOIL method to every part, you should come up with

    \frac{(k+3)(k+8)}{(k+5)(k+8)} \cdot \frac{k(k+5)}{(k-6)(k+3)}

    form here, you see that there are k+8, k+3, k+5 in both numerator and denominator, so that they would cancel out (assuming that they are not zzeros)

    so, after cancellation, only factors left are k in the numerator and k-6 in the denominator.. therefore

    \frac{k^2 +11k + 24}{k^2 + 13k + 40} \cdot \frac{k^2 + 5k}{k^2 - 3k - 18} = \frac{k}{k-6}
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