# How can I factor out a problem like this 5y(x-3)+(3-x)?

Asked by emeraldisles (1949) January 31st, 2012

I know that if the problem was 5y(x-3)+(x-3), the answer would be( x-3)(5y+1x). How do I factor out a problem in which the second set of parentheses is in reverse order(x-3)+(3-x)?

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To be precise, the factoring in your first example is not (x-3) (5y+1 x), it’s (x-3) ( 5y+1). Maybe that was just a typo on your part (but it’s the kind of typo that can lead to very wrong answers, so be careful).

In the second example, since (x-3) and (3-x) are not at all the same terms, they can’t be factored out as if they were the same. It seems to me, unless I’m missing something blindingly obvious, that the second example is already factored as much as it can be.

On the other hand, and I’m only throwing this in because we already had one typo in evidence, if the example you’re concerned about was not 5y(x-3)+(3-x) but instead 5y(x3)+(3x), then it should be clear that “x3” is the same as “3x”, and that term can be written as “3x” in both places and then factored accordingly. But I only said that because of the error that you gave in your example. It may not apply at all.

CWOTUS (26082)

Thanks. I will try it again.

emeraldisles (1949)

I’m not sure, but I think you place the negative outside, rewrite and then factor.

Ela (6496)

Yes that’s exactly what I did. I got (x-3)(5y-1) for an answer.

emeraldisles (1949)

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