Factor by grouping

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October 13th 2007, 01:31 PM
fluffy_penguin

Factor by grouping

http://i24.tinypic.com/2hncsgh.jpg

I don't understand how to do this...
October 13th 2007, 02:14 PM
topsquark

Quote:

Originally Posted by fluffy_penguin

http://i24.tinypic.com/2hncsgh.jpg

I don't understand how to do this...

Write $3a^2 + 2a - 21 = 3a^2 + (9a - 7a) - 21$

$= (3a^2 + 9a) - (7a + 21)$

$= 3a(a + 3) - 7(a + 3)$

$= (3a - 7)(a + 3)$

-Dan
October 13th 2007, 06:00 PM
fluffy_penguin

Confused

Quote:

Originally Posted by topsquark

Write $3a^2 + 2a - 21 = 3a^2 + (9a - 7a) - 21$

$= (3a^2 + 9a) - (7a + 21)$

$= 3a(a + 3) - 7(a + 3)$

$= (3a - 7)(a + 3)$

-Dan

How does

$3a^2 + 2a - 21 = 3a^2 + <b>(9a - 7a)</b> - 21$

I have no clue where the $(9a - 7a)$ came from...
October 13th 2007, 06:12 PM
Jhevon

Quote:

Originally Posted by fluffy_penguin

How does

$3a^2 + 2a - 21 = 3a^2 + <b>(9a - 7a)</b> - 21$

I have no clue where the $(9a - 7a)$ came from...

we use \bold{ } to make something bold with LaTex

topsquark noticed that 9a - 7a = 2a. how did he do that? well, he could have done it by experience, but there is a method you can do.

take the coefficient of a^2, that is 3 and multiply it by the constant term, that is, -21. the result is -63. now you want two numbers that when multiplied give -63 but when added give 2 (which is the coefficient of the middle term). 9 and -7 are such numbers. now we split the middle term into those two numbers and continue as topsquark did
October 13th 2007, 06:20 PM
fluffy_penguin

Okay thank you both of you. ^ ^
October 13th 2007, 06:29 PM
Jhevon

Quote:

Originally Posted by fluffy_penguin

Okay thank you both of you. ^ ^

you're welcome, come again :D