Psion Posted August 5, 2005 Share Posted August 5, 2005 I figured since I don't want to clutter up this board with a bunch of newly created threads, that I'd post whatever math problems I'm having in here. I suck at math, sadly. The most recent problem I've come across is this: factor each expression completely x^3 - 8y^3 So I tried doing this since: (x-y)(x^2 + xy + y^2) factored: x^3 + 2x^2y + 4xy^2 - 2x^2y - 4xy^2 - 8y^3 Which turns out to be: (x -2y)(x + 2xy + 4y) what I don't understand from this problem is how 4 gets there. Link to comment Share on other sites More sharing options...
Walden Posted August 5, 2005 Share Posted August 5, 2005 Try looking at the formula for the difference of cubes. x^3 – y^3 = (x – y) (x^2 + xy +y^2) Link to comment Share on other sites More sharing options...
DQW Posted August 5, 2005 Share Posted August 5, 2005 x^3 – 8y^3 is of the form a^3 – b^3, with a=x, b=2y. Also b^3 comes out of the factors by multiplying b*b^2. And, in this case, b=2y and b^2 = 4y^2. So, that's where the 4 comes from. PS : There's a typo in your final factorization. The last term is 4y^2, not 4y. Link to comment Share on other sites More sharing options...
Psion Posted August 5, 2005 Author Share Posted August 5, 2005 The 4 part is what is getting me. From what I figured: [math] (x - 2y)(x^2 + 2xy + 2y^2) [/math] what I don't get here is how there is suppose to be a [math]4[/math] in place of the [math]2[/math] in [math]2y^2.[/math] If i factor out 8 [math]2 * 2 * 2 = 8[/math] I don't understand the math reasoning to put a [math]4[/math] there. Link to comment Share on other sites More sharing options...
DQW Posted August 5, 2005 Share Posted August 5, 2005 8 = 2*4 Multiply out (x - 2y)(x^2 + 2xy + 2y^2) and see if you get x^3 - 8y^3. You will not ! Simply because (-2y)(2y^2) = -4y^3 instead of -8y^3, which is = (-2y)(4y^2) Link to comment Share on other sites More sharing options...
Psion Posted August 10, 2005 Author Share Posted August 10, 2005 I figured out that there is an x, y, and z variable in this word problem, but figuring it all out is somewhat confusing for me. Word Problem: The sum of three integers is 18. The third integer is four times the second, and the second integer is 6 more than the first. Find the integers. (x, y, z) Link to comment Share on other sites More sharing options...
Ducky Havok Posted August 10, 2005 Share Posted August 10, 2005 well start with x+y+z=18, z=4y, and y=6+x. The 3rd can be changed to y-6=x, then you can substitute the two back into the first to get (y-6)+y+4y=18, solve for y and just plug them back into the other equations to get the rest Link to comment Share on other sites More sharing options...
Psion Posted August 12, 2005 Author Share Posted August 12, 2005 Here are a few problems I've come across. I think I understand Gaussian Elimination but it just confuses me to some extents. Instructions: Use matrices to solve each system of equations. Each system has one solution. I'm i'm looking at: 2x + y + 3z = 3 -2x - y + z = 5 4x - 2y - 3z = 10 So I decided to turn it into [ 2 1 3|3] [-2 -1 1|5] [4 -2 2|2] So from there I decided that when doing this gaussian thing the goal is to eliminate all numbers below the first digit in the first row to 0. I decided that I didn't need to exchange rows. R=Row R1 + R2 = [0 0 4 | 8 ] (new R2) Since that was done with I decided that I now had to move to row three. So to get rid of the 4 I have to make the first row negative. R1 * -2 = [ -4 -2 -6 |-3 ] [ -4 -2 -6 |-3 ] + [ 4 -2 2|2] ---------------------- [0 -4 -4 |-1] I don't really know if I did this right, and I don't know where to go after this step. Link to comment Share on other sites More sharing options...
elfstone Posted August 12, 2005 Share Posted August 12, 2005 You want to get the matrix in this form : 1 0 0 | x 0 1 0 | y 0 0 1 | z The new R2 you did qualifies for the R3 of above if you divide it with 4. Turns out that z = 2. So we got R3 =[ 0 0 1 | 2]. The new R1 you did (slight mistake, u forgot to multiply smth there[-3 should be -6], anyway, it is 0 4 4 | 4 if multiplied with -1. This is good for our R2. R2 = R2 - 4*R3 = 0 4 0 | -4 and looks like that y = -1. R2 =[ 0 1 0 | -1]. Now, your original R1 [2 1 3 | 3] can be turned R1 = R1 - R2 - 3*R3 = [2 0 0 | -2 ] -> R1 = [1 0 0 | -1] and x is also -1. I hope that wasn't too confusing, it's been a while I did this Link to comment Share on other sites More sharing options...
Primarygun Posted August 13, 2005 Share Posted August 13, 2005 4x - 2y - 3z = 10 [4 -2 2|2] If the original problem is from your book, your answer from this approach won't be equivalent. I think you put the no. wrongly into matrix. This is the major problem I suppose. Link to comment Share on other sites More sharing options...
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