math help w/ factoring?
156986
First, I am not sure why you call this factoring. This is not factoring. \n\nThink of it this way. If each of the teams play each other once, there are G1=[N^2-N]/2 games. To see that this is true, just compute either \n\nN choose 2 = N!/[(N-2)!2!]=[N^2-N]/2 \n\nor \n\nmake a matrix with the N teams as both rows and columns. Every element of the matrix constitutes a game between two teams, one in the corresponding row and one in the corresponding column. One has to disregard the N diagonal elements, because a team cannot play itself, so there are a total of N^2-N games in the matrix, without the diagonal. Since this includes games when A plays B as well as B plays A, we have to divide by 2, to get again [N^2-N]/2.\n\nIf each of the teams play each other 2 times, there are G2=N^2-N games. To see this, just imagine the matrix of team-labelled columns and rows again, and subtract the diagonals again, so that both the games between A and B and between B and A are included, for a total of 2 games between each pair of teams. The number of elements in the matrix is N^2-N, ignoring the diagonal elements again. \n\nFor the teams to play each other 3 times, they can just play the one game schedule, and then the 2 game schedule, for a total of 3 times between each pair of teams, so:\n\nG3=G1+G2=(3/2)*[N^2-N]\n\nFor N=7, this is just 1.5*[49-7]=1.5*42=63.\n\nThere are 63 total games played.
667985
You can factor this by grouping:\n\nx^3 + 2x^2 - 9x = 18 \nx^3 + 2x^2 - 9x -18 = 0\nx^2(x + 2) + (-9)(x + 2) = 0\n(x^2 -9)(x + 2) = 0\n(x - 3)(x + 3)(x + 2) = 0\n\nx = { 3, -3, -2}\n\nFirst subtract 18 from both sides, then facter each binomial, and finally use the difference of squares.
1046507
[2x/(x+3)(x-3)]*[(x-3)/4(x)]\ncancelling 2 and x-3 we will get\n=x/(x+3)2(x)\nsimplify we will get\n=x/(2x^2+6x)
594239
ok so you have p^2 + p, you can factor out p from both parts of the expression, it will end up like this: p(p + 1).  Hope that works.
835153
.more kids homework... factoring?\n1. rewrite the middle term as the sum of two terms and then factor completely 15x^2 -21x +6\n\n3(5x^2-7x+2)\n3(5x^2-5x-2x+2)\n3(5x-2)(x-1)\n\n2. rewrite the middle term as the sum of two terms and then facto by grouping. x^2 - 14x +45 \n\nx^2-9x-5x+45\n(x-9)(x-5)\n\n3. rerewrite the middle term as the sum of two terms and then facto by grouping x^2 + 19x +90\n\nx^2-10x+9x+90\n(x+10)(x+9)\n\n4. rewrite the middle term as the sum of two terms and then facto by grouping x^2 -2x -63\n\nx^2-9x+7x-63\n(x-9)(x+7)\n\n5.solve x^2 = -6x\n\nx=-6
565133
We start by factoring the polynomial.  A little trial and error (testing the factors of 2, which is the constant term) reveals that 2 is a root.  So, we know that (x-2) is a factor.\n\nDividing the polynomial by (x-2) using synthetic division, we find the other factor is 6x^2+x-1.  This can be factored as (3x-1) (2x+1).\n\nSo, the equation, in factored form, is\n(x-2)(3x-1)(2x+1)=0\n\nSo, the roots are x=2, x=1/3, and x=-1/2.
254834
if you use quadratic formula you get the factors of \n\nx= ( ( 3 +/- square rt.(5) ) / -6 )
977979
x= 3/5\n\nIf you factor by grouping you would get\n25x^2 - 15x -15x + 9 = 0\n\nthen you would factor the first two terms and the last two terms\n5x(5x-3)   -3(5x-3)    =  0\n\nuse the terms outside the parentheses are one factor and the one inside the parentheses as the second factor\n(5x-3)(5x-3) = 0\n\nset to zero\n5x-3 = 0\n\ndo algebra\nx=3/5\n\nor you could have just noticed that that equation is a perfect square
734679
5a^2 + 3a - 2 = 0\n\ncan be factored as (5a - 2) (a + 1) = 0\nThus a = 2/5 and a = -1
858168
1. 14x^2 +21x\n\na good way to start would be to factor out any common terms:\n\n7(2x^2 + 3x)\n\n7x(2x + 3)
241036
I'm assuming you're looking for information on factoring polynomials since it's a College Algebra courese.  \nThis website is very helpful, with the steps for factoring laid out, and it also provides websites you can use if you need more information.  Good luck!\n\nhttp://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut7_factor.htm
570705
Factoring out X gives:\nX(-X^2 + X + 6)=0\nUsing the quadratic formula to factor the terms inside the brackets gives:\n-X^2 + X + 6 = (X+2)(-X+3)\nSo we have:\nX(X+2)(-X+3)=0 and the zeros are X=0, X=-2, X=3
766790
1. 5m^2 - 15mp - 2mr - 6pr <-- Impossible to factor by grouping. It needs to be either, exactly one of the signs needs to be reversed.\n\n2. 6-3x-2y+xy\n(6 - 3x) - (2y - xy)\n3(2 - x) - y(2 - x)\n(3 - y)(2 - x)
732414
Factor out 2 in the numerator.\n\n2(3x - 5)/5 - 3x\n\nMultiply the denominator by -1 and then rearrange the denominator.\n\nWe have:\n\n-1 TIMES each term inside the parentheses (5-3x).\n\nDoing this we get:\n\n-5 + 3x\n\nNOTICE: +3x = 3x.\n\nWe rerrange the denominator.\n\nWe NOW have this in the denominator: 3x - 5.\n\nLet's go back to the top.\n\n2(3x-5)/3x-5\n\nWe can CANCEL 3x-5 in the numerator and denominator.\n\nDoing that we get 2 as the answer.\n\nGuido
1053403
Write it as:\n2x^2-4x+3x-6\n=2x[x-2]+3[x-2]\n=[2x+3][x-2]\nSimple middle-term factorisation!!!
945604
1.adding-2f-7\n-6f-2f+7-7<2f-2f-9-7\n-8f<-16\ndividing by -8\nx>2\n\n2.5/8g-4>2/3g+6\nadding 4\n5/8g>2/3g+10\nmultiplying by 24\n15g>16g+240\nadding -16g\n-g>240\ndividing by -1\ng<-240\n\n3.sum incomplete\n\n4.-2 1/9d+3/5<-3 2/9d-4 2/5\n-19/9d+3/5<-29/9d-22/5\nmultiplying by45\n-95d+27<-145d-198\nadding 145d\n50d+27<-198\nadding -27\n50d<-225\ndividng by 50\nd<-5 1/2
902478
wow its been a long time.. i use a calculator to do that now.. but here goes one thing i remember...\n\nlets say we have a polynomial like this\n\nx^2 -3x-10  and we need to factor it..\n\nwe take a look at what factors of -10 we could use that would add up to -3  in this case there are two way to get -10 (-2*5) or (2*-5)  if we add the two sets together one gives us -3 and the other gives us 3 so we can throw out the -2*5 terms.\n\nSo now we lay out the factors.... we need two x's to get x^2 so we need two x's in the first terms (remember FOIL?)\n\nlay the terms out so they give us the correct values\n\n(x+2)(x-5)  now we can use FOIL (firs, outer, inner, last) to check ourselves...\nX*X = x^2\nX*-5 = -5X\n2*x =2x\n2*-5=-10\n\nso we get [x^2] +[-5x] +[2x] +[-10]\nor x^2 -5x+2x-10\ncombine like terms now\n-5x+2x =-3x\n\nwhich gives us\nx^2-3x-10\n\nhope that helps