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Thread: I need help with more factorizaing...

  1. #1
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    I need help with more factorizaing...

    factorization is not my thing
    anywayz here are the two questions

    x^2 +3x(x+3y)+2(x+3y)^2

    and

    3(2x-y)^2 +2y(2x-y)-5y^2

    steps would be nice
    thank you!
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  2. #2
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    Quote Originally Posted by xo-sherry View Post
    factorization is not my thing
    anywayz here are the two questions

    x^2 +3x(x+3y)+2(x+3y)^2

    and

    3(2x-y)^2 +2y(2x-y)-5y^2

    steps would be nice
    thank you!
    Let $\displaystyle Y = x + 3y$.

    So the first becomes

    $\displaystyle x^2 + 3xY + 2Y^2$.

    $\displaystyle = x^2 + xY + 2xY + 2Y^2$

    $\displaystyle = x(x + Y) + 2Y(x + Y)$

    $\displaystyle = (x + Y)(x + 2Y)$

    $\displaystyle = (x + x + 3y)[x + 2(x + 3y)]$

    $\displaystyle = (2x + 3y)(x + 2x + 6y)$

    $\displaystyle = (2x + 3y)(3x + 6y)$

    $\displaystyle = 3(2x + 3y)(x + y)$.


    For the second, let $\displaystyle X = 2x - y$, so it becomes

    $\displaystyle 3X^2 + 2yX - 5y^2$


    $\displaystyle 3 \times -5 = -15$ so we need two numbers that add to become $\displaystyle 2$ and multiply to become $\displaystyle -15$. They are $\displaystyle 5$ and $\displaystyle -3$. So we break up the middle term into $\displaystyle 5yX - 3yX$.


    $\displaystyle 3X^2 + 2yX - 5y^2 = 3X^2 + 5yX - 3yX - 5y^2$

    $\displaystyle = X(3X + 5y) - y(3X + 5y)$

    $\displaystyle = (3X + 5y)(X - y)$

    $\displaystyle = [3(2x - y) + 5y](2x - y - y)$

    $\displaystyle = (6x - 3y + 5y)(2x - 2y)$

    $\displaystyle = 2(6x + 2y)(x - y)$

    $\displaystyle = 4(3x + y)(x - y)$.
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