Drunk math

I may have had a few drinks. What I mean by that is that I had a few drinks about 5 hours ago and still feel kind of wasted. Do I really have nothing better to do on a saturday? I blame summer and keeping my mind inactive

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(x+1)(x-2)(x+4): All possible combinations to make x^3 = x*x*x, all possible combinations to make x^2 = x*x*4 + x*-2*x + 1 * x * x = 3x^2; all possible combinations to make x = x*-2*4 + 1*x*4 + 1*-2*x = -6x, all possible combinations to make zero degree term = 1*-2*4 = -8. => (x+1)(x-2)(x+4) = x^3 + 3x^2 - 6x -8. That was easy

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Whenever you multiply 3 binomials that all take the form (x + a)(x + b)(x + c) (which can easily be extended), you get x^3 + (a + b + c)x^2 + (ab + ac + bc)x^ + a*b*c. So if you had, x^3 + 3x^2 -6x -8, how can you factor that? well we assume they take that form, then a + b + c = 3, ab + ac + bc = -6, and a * b * c = -8.

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From the third equation, we can see that either a,b,c are all negative or just one is negative. Since the first equation, if you add 3 negative numbers then you get a negative number so only 1 is negative. Then we think of 3 numbers that multiply be -8. The factors of -8 are 2, 2, and 2. combinations are 2, 4, 1; 2, 2, 2; and 8, 1, 1. To make 3, we would take 1 + 4 - 2. To check; 1*4 + 1 * -2 + 4 * -2 = 4 -2 -8 = -6

Comments

[quote name='sinharvest24' post='3753094' date='Jul 4 2011, 06:42 AM']If anyone in here can follow/understand what Zetta did, i applaud you.[/quote]
I got the first one immediately. I haven't read the rest.
This shit pops into my head all the time, without booze or drugs. That's probably... bad.
 
I remember the last time you have a few drinks. XD You sat on your Micro SD Card, did a nice jump over a coffee table, and sang the Pokemon theme song. (BTW: Your Internet is in the mail, I sent it via Priority shipping.)
 

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