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    Math People

    How do I factor by grouping:

    x^3 - x +5x^2 - 5

    This shit blows, any help would be phenomenal.

    #2
    Math is for herbs. Drop the fuck out.

    Comment


      #3
      this is exactly why i haven't gone to college yet. i haven't taken a math class in 3 years and it's been so nice.

      sorry i couldn't help ya though

      Comment


        #4
        You can't smoke weed and be good at math.

        Comment


          #5
          When are you ever going to need to know that in real life?

          Comment


            #6
            Originally posted by Charlie Hustles
            Math is for herbs. Drop the fuck out.
            kthx

            Comment


              #7
              For real, though, I used to be really good at math and I know that I knew how to do that, but after a good 6 years of never having to use it once, I forgot.

              Comment


                #8
                Originally posted by ian rogers | phc
                You can't smoke weed and be good at math.
                thats not true, I have one friend who does smack, is an economist major and was awarded some second best scholar @ OSU WHILE having dope/water mixed in a nose spray bottle/doing it on stage.

                another friend who is an electrical engineer and the most efficient math student I have ever seen.(forgot to add this) but he has smoked a shit load of pot since we were like 15, and does the smack occasionally as well.

                now that I think about it, H is pretty big in this town. its the sux =(

                you ever hear of those big professionals that get caught being opiate addicts(doctors, lawyers, etc) well, they all come from here

                You have to be able to function in everyday life on drugs here in Oregon, there is just nothing else to do.

                Comment


                  #9
                  Re: Math People

                  Originally posted by Coopnasty
                  How do I factor by grouping:

                  x^3 - x +5x^2 - 5

                  This shit blows, any help would be phenomenal.
                  x(x^2 - 1) + 5(x^2 - 1)

                  =

                  (x + 5) (x^2 - 1)

                  =

                  (x + 5) (x + 1) (x - 1)




                  Correct?

                  Comment


                    #10
                    X^2(x+5) + -1(x+5)

                    Comment


                      #11
                      Re: Math People

                      Originally posted by BCarr4328
                      Originally posted by Coopnasty
                      How do I factor by grouping:

                      x^3 - x +5x^2 - 5

                      This shit blows, any help would be phenomenal.
                      x(x^2 - 1) + 5(x^2 - 1)

                      =

                      (x + 5) (x^2 - 1)

                      =

                      (x + 5) (x + 1) (x - 1)




                      Correct?
                      Actually you're correct, thank you!

                      Comment


                        #12
                        You basically seperate the polynomial into two parts and pull out the greatest common factor. So you have....

                        X^3-X + 5x^2 - 5

                        Then do each seperately then combine

                        X(X^2-1) + 5(X^2-1)

                        Now the expression in the paretheses are the same so you can combine (I think it is the reverse of the distributive property)

                        (X+5)(X^2-1)

                        X^2-1 is a perfect square so the answer is...

                        (X+5)(X+1)(X-1)

                        Comment


                          #13
                          Fucking barr.

                          Fuck math, fuck factors by grouping, and fuck my google search.

                          Comment


                            #14
                            Re: Math People

                            Originally posted by Coopnasty
                            Originally posted by BCarr4328
                            Originally posted by Coopnasty
                            How do I factor by grouping:

                            x^3 - x +5x^2 - 5

                            This shit blows, any help would be phenomenal.
                            x(x^2 - 1) + 5(x^2 - 1)

                            =

                            (x + 5) (x^2 - 1)

                            =

                            (x + 5) (x + 1) (x - 1)




                            Correct?
                            Actually you're correct, thank you!
                            Dang Straight

                            Comment


                              #15
                              x^3 - x + 5x^2 - 5

                              x (x^2 - 1) + 5 (x^2 - 1)

                              (x^2 - 1) (x + 5)

                              EDIT:
                              nm... got beat to the answer... and i guess i didnt factor it far enough

                              Comment


                                #16
                                Originally posted by RobbedBlind
                                x^3 - x + 5x^2 - 5

                                x (x^2 - 1) + 5 (x^2 - 1)

                                (x^2 - 1) (x + 5)
                                ok he obviously already got it.

                                i beat you all, Bitches

                                Comment


                                  #17
                                  ok, you need to put it in order first

                                  x^3 + 5x^2 - x - 5

                                  no you have an x^3, x^2 and x variable terms
                                  the common denominator of these is x and x^2 so start there

                                  (x^2.....)(x....)

                                  look at what you're left with....

                                  "+5x^2" shows you that the x^2 term must be mulltiplied in the parenthasis with the x....

                                  while the "-5" must be multiplied by something... since we already have a 5 then it must leave -1 in the "x^2..." parenthesis.

                                  (x^2 - 1)(x + 5)
                                  Worked

                                  Comment


                                    #18
                                    Originally posted by Charlie Hustles
                                    When are you ever going to need to know that in real life?

                                    Comment


                                      #19
                                      Originally posted by Charlie Hustles
                                      Originally posted by Charlie Hustles
                                      When are you ever going to need to know that in real life?
                                      Oh,at the grocery store of course.

                                      Comment


                                        #20
                                        Originally posted by Charlie Hustles
                                        Originally posted by Charlie Hustles
                                        When are you ever going to need to know that in real life?
                                        when dealing with a function that describes some sort of observable where the outcome is already known.

                                        something that has an exponential growth which is determined by more than one term can be solved backwards to yield the rate of growth.

                                        it's ironic that it's "exponentially" useful

                                        it's simply a separation of terms, the same method would be used if the function contained more than one variable, this same method would be a sound way to extract physical data and used to make future predictions based off of previous observables(which dictated the actual function).
                                        Worked

                                        Comment


                                          #21
                                          Originally posted by Guest!mator
                                          Originally posted by Charlie Hustles
                                          Originally posted by Charlie Hustles
                                          When are you ever going to need to know that in real life?
                                          when dealing with a function that describes some sort of observable where the outcome is already known.

                                          something that has an exponential growth which is determined by more than one term can be solved backwards to yield the rate of growth.

                                          it's ironic that it's "exponentially" useful

                                          it's simply a separation of terms, the same method would be used if the function contained more than one variable, this same method would be a sound way to extract physical data and used to make future predictions based off of previous observables(which dictated the actual function).
                                          Is this a joke?

                                          Comment


                                            #22
                                            no, that's when you would use it in real life. people really do this because math provides reliable interpretations and sound predictions of data.
                                            Worked

                                            Comment


                                              #23
                                              sometimes i feel like people ask me a question, not expecting a) me to take it seriously or b) a real answer.
                                              Worked

                                              Comment


                                                #24
                                                Your sig is incredible

                                                Comment


                                                  #25
                                                  Originally posted by Guest!mator
                                                  no, that's when you would use it in real life. people really do this because math provides reliable interpretations and sound predictions of data.
                                                  Lame, he was asking it in the sense of saying no one uses that really. Not actually asking i thought.
                                                  And you just said how to use it not actually saying where like "At business' statistician groups."
                                                  And you explained it so confusingly, if he cant do the problem then i'm sure he will need a more understandable definition. I Had no idea what your saying.

                                                  i just felt like yelling at someone forgive me?

                                                  Comment


                                                    #26
                                                    Why does that matter?

                                                    Comment


                                                      #27
                                                      i have to use technical terms because that's the best way to describe what's going on. i'm not that good at describing a real life example because theyre generally pretty complex. but if you're gonna work at mcdonalds, then no you won't need it.
                                                      Worked

                                                      Comment


                                                        #28
                                                        But, what would you need it for?

                                                        Comment


                                                          #29
                                                          Originally posted by Charlie Hustles
                                                          But, what would you need it for?
                                                          Stealing HV 30's and stuff of that nature obviously

                                                          Comment


                                                            #30
                                                            Bahahaha, but seriously . . .

                                                            Comment


                                                              #31
                                                              You need to know how to separate the variables...in simple terms, how else would you find out what the variable X would be if you knew that function was equivalent to zero. It helps you understand how all of math is related to each other and you can use these properties with many, if not all, math circumstances.

                                                              I can explain when calculus can be used...

                                                              Here's a small calculus example using inflection points and derivatives and a real life application + brief description of what a derivative is...this is probably gonna be over some people's heads.

                                                              derivatives- a function that describes a plot on a graph(any function, such as how many girls versus how many boys came to the party)... when you take the derivative of that function it will give you an equation for the slope of that function at that specific point(usually defined by the variable X).

                                                              Now for a real life example that i had to do in the lab recently:

                                                              Gas Chromatography is a fancy name for separation of compounds via thermodynamics affecting retention time of a gas within a column lined with polar stationary phase, which will allow the polar compounds to "take longer" to come out of the column than the non-polar compounds because they have no interaction with the column, unlike the polar compounds.

                                                              Anyways, a carrier gas, Helium in tnis case, is used to carry these compounds that need to be separated, down the column. In order to get a proper separation and adequate resolution on the gas chromatograph report, the carrier gas speed must be at "optimum flow rate". The optimum flow rate is determined, in a complex graph, but easily understood by this simple explanation: at slower carrier gas speeds there is lack of resolution because there becomes something called eddy diffusion which causes the gas to be retained too long, however u can make the eddy diffusion a minimum if you speed it up, however that in turn makes the resolution go down because if the carrier gas flow rate is too high then you lose resolution due to mass transfer of the gas solute. SO the graph basically has an inflection point(meaning the graph changed directions and looks like a rollercoaster hill/the dip- we are most concerned with this "dip" because that is the maximum resolution and the "optimum flow rate")

                                                              because we can take the graph and create a function to represent that graphic, we can then find the exact point where the graph is a minimum(the dip in the graph will have a zero slope at it's inflection point, this means that at that one spot/the optimum flow rate/ the slope would be equal to zero because the function is changing directions)

                                                              take the function for the graph and find it's derivative(which is the slope function now) and set it equal to zero and solve for the variable(the variable determines where the exact minimum flow rate is)

                                                              This is just the tip of why calculus is so important...because a lot of complex and crazy things are described by calculus and higher math. Computer graphics use intense wave equations to represent the contour and shading of objects, those are purely math functions in 3-d! it's crazy stuff.

                                                              to relate that explanation back to the first function. If the graph had 2 inflection points then the slope function would yield some sort of multiple variable scenario, such as the original X^3..... function we saw in the begining. separating and solving for the multiple variables would give the multiple inflection points on the graph(which can be very useful).
                                                              Worked

                                                              Comment


                                                                #32
                                                                factor by grouping is easy

                                                                Comment


                                                                  #33
                                                                  Originally posted by Guest!mator
                                                                  You need to know how to separate the variables...in simple terms, how else would you find out what the variable X would be if you knew that function was equivalent to zero. It helps you understand how all of math is related to each other and you can use these properties with many, if not all, math circumstances.

                                                                  I can explain when calculus can be used...

                                                                  Here's a small calculus example using inflection points and derivatives and a real life application + brief description of what a derivative is...this is probably gonna be over some people's heads.

                                                                  derivatives- a function that describes a plot on a graph(any function, such as how many girls versus how many boys came to the party)... when you take the derivative of that function it will give you an equation for the slope of that function at that specific point(usually defined by the variable X).

                                                                  Now for a real life example that i had to do in the lab recently:

                                                                  Gas Chromatography is a fancy name for separation of compounds via thermodynamics affecting retention time of a gas within a column lined with polar stationary phase, which will allow the polar compounds to "take longer" to come out of the column than the non-polar compounds because they have no interaction with the column, unlike the polar compounds.

                                                                  Anyways, a carrier gas, Helium in tnis case, is used to carry these compounds that need to be separated, down the column. In order to get a proper separation and adequate resolution on the gas chromatograph report, the carrier gas speed must be at "optimum flow rate". The optimum flow rate is determined, in a complex graph, but easily understood by this simple explanation: at slower carrier gas speeds there is lack of resolution because there becomes something called eddy diffusion which causes the gas to be retained too long, however u can make the eddy diffusion a minimum if you speed it up, however that in turn makes the resolution go down because if the carrier gas flow rate is too high then you lose resolution due to mass transfer of the gas solute. SO the graph basically has an inflection point(meaning the graph changed directions and looks like a rollercoaster hill/the dip- we are most concerned with this "dip" because that is the maximum resolution and the "optimum flow rate")

                                                                  because we can take the graph and create a function to represent that graphic, we can then find the exact point where the graph is a minimum(the dip in the graph will have a zero slope at it's inflection point, this means that at that one spot/the optimum flow rate/ the slope would be equal to zero because the function is changing directions)

                                                                  take the function for the graph and find it's derivative(which is the slope function now) and set it equal to zero and solve for the variable(the variable determines where the exact minimum flow rate is)

                                                                  This is just the tip of why calculus is so important...because a lot of complex and crazy things are described by calculus and higher math. Computer graphics use intense wave equations to represent the contour and shading of objects, those are purely math functions in 3-d! it's crazy stuff.

                                                                  to relate that explanation back to the first function. If the graph had 2 inflection points then the slope function would yield some sort of multiple variable scenario, such as the original X^3..... function we saw in the begining. separating and solving for the multiple variables would give the multiple inflection points on the graph(which can be very useful).
                                                                  wtf was the point of this?

                                                                  Comment


                                                                    #34
                                                                    Originally posted by Charlie Hustles
                                                                    But, what would you need it for?
                                                                    Sometimes, when people grow up, they do "science" and "engineering." You'll learn more about it at Career Day.

                                                                    Comment


                                                                      #35
                                                                      Originally posted by blowakilo
                                                                      Originally posted by ian rogers | phc
                                                                      You can't smoke weed and be good at math.
                                                                      thats not true, I have one friend who does smack, is an economist major and was awarded some second best scholar @ OSU WHILE having dope/water mixed in a nose spray bottle/doing it on stage.

                                                                      another friend who is an electrical engineer and the most efficient math student I have ever seen.(forgot to add this) but he has smoked a shit load of pot since we were like 15, and does the smack occasionally as well.

                                                                      sounds like a winner to me

                                                                      Comment


                                                                        #36
                                                                        Originally posted by rev. the mike
                                                                        Originally posted by Charlie Hustles
                                                                        But, what would you need it for?
                                                                        Sometimes, when people grow up, they do "science" and "engineering." You'll learn more about it at Career Day.
                                                                        But.. but... what do you need those for?

                                                                        Comment

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