How do i do factoring?
470305
Factoring is the "opposite" of FOILing.\n\nIf you had x^2 (which is x squared)\nx^2 + 5x + 6\n\nYou ask yourself:  What multiplies to 6?  (Write all the choices including negatives)\nThen, out of those choices, which one (and only one will work) adds to 5.  \n\nSo... 2 x 3 = 6  and -2 x -3 = 6\n2 + 3 = 5\n-2 + -3 = -5 \n\nSo, 2 and 3 add to 5 and mult. to 6.\n\nThen you write it in parenthesis keeping the signs.\n\nYou get (x + 2)(x + 3)\n\nHope that helps.  Good luck.
890098
I enjoy algebra, and this is easily my favorite type of problem.  So I hope you, too, one day enjoy them for their inner beauty.\n\nFor now, let's tackle this "factoring problem."  We may wish to use the method of "synthetic division" which is arguably easier than "long division of polynomials."\n\nThink back to grade school where you learned that 6 = 2 * 3.  This simple statement enabled you to 'deconstruct' the number 6 as either "two times three" or "three times two".  Even if you forgot one of the 'factors' you could conjure it up again quickly by asking yourself, "What is six divided by two?" for example.\n\nIn problem (a), you are essentially asking, "What are the coefficients of P(x) if the only factor we know is (x-1)?"\n\nLet's set up synthetic division (see link below), by placing all the coefficients of P(x), and using 1 (the zero implied by (x-1)) as the divisor:\n\n1 ||  a  b  c  d\n___________\n.....a\n\n1 ||  a  b  c  d\n_____a_____\n.....a (a+b)\n\n1 ||  a  b        c       d\n_____a___(a+b)____\n.....a (a+b) (a+b+c)\n\n1 ||  a  b        c                d\n_____a___(a+b)___(a+b+c)_\n.....a (a+b) (a+b+c) (a+b+c+d)\n\nNote that this last term must equal 0 since it is assumed that (x-1) is a factor ... thus, the "remainder" term must equal zero!\n\nThere's the result you were after:  a+b+c+d = 0\n\nHope this helps you tackle the others!
555590
Just multiply:\n(X/3)(3/4 x) -(x/3)(3/5) + (3/4)(3/4x) - (3/4)(3/5)\n\nand simplify from there
208169
Factoring, I guess, but it really depends on the type of polynomial question. Can you give a sample?
159156
factoring, then:\n\n(2x-3)(x+1)=0\n\nyou can assume that\n\n2x-3 = 0, and\n\nx+1 = 0\n\nSolve for first x,\n\n2x=3 , x=3/2\n\nsecond x,\nx= -1\n\nThe sum would be 3/2 - 1, or 3/2 - 2/2 =1/2
1119411
1. Add 3 to both sides to set equal to zero, then factor to get something that adds to three and multiplies to 28---7 and -4\n(x-4)(x+7) = 0\nx=4, 7\n\n2. Add 4 to both sides to set equal to zero, then factor to get something that adds to 2 and multiplies to -15---5 and -3\n(x+5)(x-3) = 0\nx = -5, 3
777394
factor out 4b\n\n4b(b-2)= 0\n\nb = 2, 0\n\ngoodluck ^_^
803895
To factor a number means to break it up into numbers that can be multiplied together to get the original number.
91753
factorize\nx^2 + x = x(x+1)\n\nfactorize\nx^2 - 1 = .... (you should know)\n\n2 = 2 (!!!)\nTake the lcm of the three denominators (each different factor (=parenthesys) once and with the highest power)\n\nIt is 2x(x+1)(...)   (fill the dots)\n\n.....\n\nnow if you can't go on, you need to open your book again and study, or pay more attention at school at your math classes, and ask immediately for further explanation to your teacher when you feel you've not understood.\n\nGood work and good study
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
713027
Just try out small numbers. 1, 2, 3, 4, 6, 8 all work. Once you start doing anything higher, the second factor becomes smaller again, so the others are 96/8, 96/6, 96/4, 96/3, 96/2, 96/1.\nIn all: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96.
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
779993
if a is a zero x-a is a factor\n1)\nso factors are (x+2) ,x (x-5)(x-6)\n\nmultiply them and you get\nf(x) = x(x+2)(x-5)(x-6) \n\nmultiply them and you get leading coefficient 1\n\n2)\n factors are (x-(1-sqrt(5)) and (x-(1+sqrt(5))\n\nso (x-1-sqrt(5))(x-1+ sqrt(5)) (a+b a-b form )\n= (x-1)^2 - 5  \n= x^2-2x-4
813960
To get a factor tree, start with picking a couple factors of 1200 (say 30 and 40).  Draw them as two branches under 1200.  Now continue doing that for each number (30 = 5 x 6, 40 = 4 x 10).  Keep branching until you get to a prime number.  When you read all the "leaves" you'll have your prime factorization.\n\n... 1200\n..../ .... &#92;\n.30 .....40\n./ .&#92; .... /. &#92;\n5  6... 4 10\n.. / &#92; ../ &#92; . / &#92;\n..2 3 2 2 2 5\n\n1,200 = 2 x 2 x 2 x 2 x 3 x 5 x 5\n\nNote: it doesn't matter how you pick your factors, since you'll still get to the same leaves.  The example above is just one way to create a factor tree for 1,200.  You could have started with 12 x 100, for example.  Just pick whatever first comes to mind.
1174486
First look at the highest powers of a and b which in this case is "squared" and the product 6ab is present, so factoring will likely involve (a+b)^2\nIn fact 3a^2+6ab+3b^2 = 3(a^2+2ab+b^2) = 3(a+b)^2\nSo your problem reduces to:\n\n3(a+b)^2 +7(a+b) - 6\n\nIf we choose to factor -6 as -2 x 3 then the it is easy to factor, and the solution is:\n\n(3(a+b) -2)((a+b)+3)
930308
x=[2+/-rt(4+60)]/2\n=[2+/-8]/2\n=-3 or 5\nso the factors are\n(x+3)(x-5)
932672
Add 20 to both sides to get a zero on the products side\n6x^2 + 19x - 10 + 20 = -20 + 20\n6x^2 + 19x  + 10 = 0\nTo factor it the basic formula is:\n(_x + _)(_x + _)\n\nNow you need to figure out the factors\nTo get six your possible factors are 1 and 6 or 2 and 3\nTo get ten your possible factors are 1 and 10, or 2 and 5\n\nTo get the middle term of 19x you'll need to play around with the numbers and orders until when you FOIL it to check yourself you get 6x^2 + 19x  + 10 = 0\n\nIt just so happens to be\n(3x + 2)(2x + 5) = 0\nAnd to check yourself:\n6x^2 + 2(2x) + 5(3x) + 10 = 0 --> 6x^2 + 4x + 15x + 10 = 0 --> 6x^2 + 19x + 10 = 0\n\nOkay now lets simplify it to find the x values\n3x +2 = 0 --> 3x = -2 --> x = -2/3\n2x + 5 = 0 --> 2x = - 5 --> x = -5/2 --> x = -2.5\nSo x can be -2/3 and - 2.5