# Factoring by group/grouping

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• Apr 26th 2010, 05:20 AM
yorkey
Factoring by group/grouping
Hi, I have a funny little question that is:

jam - pam - jim + pim

My workings are like this:
jam - pam - jim + pim
= (jam - pam) + (jim - pim)
= am (j - p) + im (j - p)
= (am + im)(j - p)

Can anyone tell me if this is correct? Thank you :)
• Apr 26th 2010, 05:31 AM
Soroban
Hello, yorkey!

You missed on a minus-sign and . . .

Quote:

Factor: . $jam - pam - jim + pim$

My workings are like this:

$jam - pam - jim + pim$

. . $=\; (jam - pam) \:{\color{red}-}\: (jim - pim)$

. . $=\; am (j - p) \:{\color{red}-}\;im (j - p)$

. . $=\; (am \:{\color{red}-}\: im)(j - p)$

. . $=\;{\color{red}m}(a-i)(j-p)$

• Apr 26th 2010, 05:47 AM
yorkey
Thank you for the quick reply! It's amazing how helpful some people, like yourself, are.

But I don't understand the 1st line: If I want to change the sign inside the bracket from a plus sign to a negative sign, shouldn't I then change the sign outside of the bracket from a negative sign to a plus sign?
• May 10th 2010, 02:54 AM
yorkey
Hi,

I'm sorry to reply again, but the above question is still bothering me.

If I have this part of a sum:

Quote:

- (jim + pim)
, won't that change to

Quote:

+ (jim - pim)
I swap the signs around?
• May 10th 2010, 10:21 PM
oregon88
Yes an example of this is

-1(2+3)
You would just factor -1 into the problem changing them both (-2-3)