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 :)

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- Apr 26th 2010, 04:20 AMyorkeyFactoring 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, 04:31 AMSoroban
Hello, yorkey!

You missed on a minus-sign and . . .

Quote:

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

My workings are like this:

$\displaystyle jam - pam - jim + pim$

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

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

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

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

- Apr 26th 2010, 04:47 AMyorkey
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, 01:54 AMyorkey
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)

Quote:

+ (jim - pim)

- May 10th 2010, 09:21 PMoregon88
Yes an example of this is

-1(2+3)

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