# Common Factors

Printable View

• May 4th 2010, 10:21 PM
larry21
Common 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, 10:23 PM
pickslides
$xy + 12 + 4x + 3y$

$xy + 3y + 4x+ 12$

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

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

Do you follow?
• May 4th 2010, 10:33 PM
larry21
Quote:

Originally Posted by pickslides
$xy + 12 + 4x + 3y$

$xy + 3y + 4x+ 12$

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

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

Do you follow?

Do you always have to rearrange the expression before solving?

Can't it also say x(y+4)+3(4+y) in the 3rd step?

And how would you know to group which ones together?
• May 5th 2010, 04:45 AM
pickslides
Quote:

Originally Posted by larry21
Do you always have to rearrange the expression before solving?

Only so you can take out a common factor.

Quote:

Originally Posted by larry21
Can't it also say x(y+4)+3(4+y) in the 3rd step?

Yep, addition commutes.

Quote:

Originally Posted by larry21

And how would you know to group which ones together?

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.