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x(x-1)(x+1)
do I multiply the brackets by eachother first, or each bracket by x?
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Multiplication is commutative, that is 6 x 7 = 7 x 6.
When you have three terms to be multiplied, first multiply the first and the second and the result by the third.It wouldn't make a difference if you multiply the second and the third and then multiply the result by the first.
x(x-1)(x+1) = (x²-x)(x+1) = x³ + x² - x² - x = x³ - x.
(x-1)(x+1)(x) = (x² - 1)(x) = x³ - x
As you can see, the results are the same in both the cases.
When you have a(x-y), you'd have to multiply both the terms inside the bracket by a, you get ax - ay.
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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or you simply can see a formula (a-b)(a+b)=a² - b² ; as a result (x-1)(x+1)=x²-1²
so x(x-1)(x+1)=x(x²-1²)=x³-x
Last edited by Rose-Red (2006-01-05 21:54:54)
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Thankyou, I was drunk
Won't happen again.
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Maybe someone could simplify this?
again, I need the whole solution, because the answer is at the end of the book
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I'd start with x/x^2, firstly I'd "reciprocal" which basically means you take the value with the power (x^2) and move it to the top, when you do this you MUST change the sign of the power:
x / x^2 = x * x^-2
I'd then do the same with the next part of the sum, giving you b - ax * x^-3. Now all that is left is
(x * x^-2) + (b - ax * x^-3), since none of the values share similar powers, I think this is as far as you can go. (Don't take my word for it, I'm new to this too).
Last edited by rickyoswaldiow (2006-01-06 10:06:35)
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well the answer is
I also tried to do that but in quite different way. Still I couldnt get the right answer. That's what I got
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Sounds like a job for Crazy Holmes
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I think the book's answer isn't correct. We can't eliminate a.
I got the same as you.
IPBLE: Increasing Performance By Lowering Expectations.
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yeah, maybe you're right.
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I've just read this "crazy holmes" thing.
Ha, ha, ha, that was VERY VERY... VERY
### ### ### ###
# # # # # # #
# ## # # # # # #
# # # # # # # #
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IPBLE: Increasing Performance By Lowering Expectations.
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Actually it was BAD, but I'll forgive you if you promise that you will
NEVER
NEVER
NEVER
do this again
much
IPBLE: Increasing Performance By Lowering Expectations.
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appologise, grassy homes.
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Very funny. First krassi_holmz becomes crazy holmes, finally, grassy homes. I really hope krassi_holmz doesn't take offence at that. Maybe, rickyoswaldiow is just being funny. But a word of caution to rickyoswaldiow. People don't like their names (or usernames) spelt or pronouned wrongly.
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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Yes, wildrocky, yes.
IPBLE: Increasing Performance By Lowering Expectations.
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No offence intended. I play online FPS games and my language in that would get me banned in a flash here. You **...
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I'm a bit late here, but I agree with everyone else.
I'd just scale up the left fraction by x to get a common denominator of x³, and then they can easily be added together.
Why did the vector cross the road?
It wanted to be normal.
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Yes, but the "a" cannot be removed.
IPBLE: Increasing Performance By Lowering Expectations.
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