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 :D

thank you!

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- Mar 21st 2009, 09:44 PMxo-sherryI 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 :D

thank you! - Mar 21st 2009, 10:44 PMProve It
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 $\displaystyle 5yX - 3yX$.

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

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

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

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

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

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

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