Factor by grouping

xy + 12 + 4x + 3y = (x+3)(4+y)<-------txtbk ans

x² + x - xy - y = (x+1)(x-y)<--------txtbk ans

Please help. I'm confused how to arrive to the answer. Can someone show me the steps? Thank you!

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- May 4th 2010, 09:21 PMlarry21Common Factors
Factor by grouping

xy + 12 + 4x + 3y = (x+3)(4+y)<-------txtbk ans

x² + x - xy - y = (x+1)(x-y)<--------txtbk ans

Please help. I'm confused how to arrive to the answer. Can someone show me the steps? Thank you! - May 4th 2010, 09:23 PMpickslides
$\displaystyle xy + 12 + 4x + 3y$

$\displaystyle xy + 3y + 4x+ 12 $

$\displaystyle y(x + 3) + 4(x+ 3) $

$\displaystyle (x + 3)(y+4) $

Do you follow? - May 4th 2010, 09:33 PMlarry21
- May 5th 2010, 03:45 AMpickslides
Only so you can take out a common factor.

Yep, addition commutes.

You try to find groups where there is a common factor that once taken out gives the same inside each bracket. Sometimes might take a little trial and error. Perseverance will be your friend.