I need help with more factorizaing...

**xo-sherry** · March 21st 2009, 09:44 PM

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!

**Prove It** · March 21st 2009, 10:44 PM

Originally Posted by xo-sherry

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 $Y = x + 3y$ .

So the first becomes

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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