What rule do I break here?  TOPIC_SOLVED

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What rule do I break here?

Postby Jherek2 on Mon Oct 29, 2012 10:07 pm

I'm factoring the difference of two cubes: mx3+my3-x-y,
I get this far
m(x+y)(x2+xy+y2)-x-y
and then I do this
m(x+y)(x2+xy+y2)-1(x+y) to get (m-1)(x+y)(x2+xy+y2) which is wrong.

I've worked out what I should have done (I think!),
(x+y)m(x2+xy+y2)-x-y
(x+y)(mx2+mxy+my2)-x-y
(x+y)(mx2+mxy+my2)-1(x+y)
(x+y)(mx2+mxy+my2-1)

The worrying thing is, I don't know WHY I can't do what I did when I went wrong :oops: ! So how do I recognise when I'm going to do it again? :?:
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  TOPIC_SOLVED

Postby stapel_eliz on Mon Oct 29, 2012 11:07 pm

Jherek2 wrote:I'm factoring the difference of two cubes: mx3+my3-x-y,
I get this far
m(x+y)(x2+xy+y2)-x-y
and then I do this
m(x+y)(x2+xy+y2)-1(x+y) to get (m-1)(x+y)(x2+xy+y2) which is wrong....

The worrying thing is, I don't know WHY I can't do what I did when I went wrong

Try highlighting things in excruciating clarity... 8-)

. . . . .m(x + y)(x2 + xy + y2) - x - y

. . . . .m(x + y)(x2 + xy + y2) - 1(x + y)

Can you see now that the only thing that factors out front is the (x + y), leaving the m(x2 + xy + y2) - 1 to be combined and simplified inside the parentheses? :wink:
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Re: What rule do I break here?

Postby Jherek2 on Tue Oct 30, 2012 7:18 pm

Thank you for the explanation :D . When someone shows me something I'm stuck on, it always seems so easy! I need to slow down a bit and get more practice!
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