Here is a fun math problem about factoring!!  See if u can figure this one out!?
84658
i got it!\n\nit's 3(2c+7)(c-4d)(c+d)!!!!\n\n6c^3 + 21c^2 - 18dc^2 - 63cd - 24cd^2 - 84d^2\nfirst we take out factor 3:\n3(2c^3 + 7c^2 - 6dc^2 - 21cd - 8cd^2 - 28d^2)\nthen we take out (2c+7) from each pain inside the parenthesses:\n3[(2c+7)c^2 - (2c+7)3cd - (2c+7)4d^2)]\nwe take (2c+7) out:\n3(2c+7)[c^2 - 3cd - 4d^2]\nwe factor [c^2 - 3cd - 4d^2] trinomially:\nc^2 - 3cd - 4d^2 = (c-4d)(c+d)\nthen we get:\n3(2c+7)(c-4d)(c+d)\n\nam i right??\n\n:) yeah i didn't believe myself...\nbut if u really write down all your factors, it's much more easier to take them out. and it WAS fun :))
87985
The expression was not set equal to anything.\n\nIf you're doing simple factoring in algebra class, the \nsimplest factors would be (b+4) (b+3).\n\nIf the expression was set to equal 0, solve for b, then your answer set would include -4 and -3.\n\nHope this helps!
264056
k=12, assuming you meant 4(k+2) - 2(k-1) = 30, otherwise it won't work out.
796585
1.factor\n(2x+1)(x-4)=0\n2.set it to 0\n2x+1=0\nor\nx-4=0\n3.solve\n{x=-1/2 or x=4} <=answer
945958
(2n-1) * (2n-2) * (2n-3) * (2n-4)... * 3 * 2 * 1\n\nJust like any other factorial.
136789
That's an extremely good question. \n\nThe answer is simple. Before that, i will explain what your question means to those who don't get. Well, it means that we must get 24 using 1,3,4 and 6 any number of times however involving operations +,-,* and / only once. \n\n  Then, the answer is very simple :\n\n  {[(6*6)/(3-1)]+[6]}  =  24\n\n---------------------------------------------------------\nWOKRING :\n\n {[(6*6)/(3-1)]+[6]}   =  {[(36)/(2)]+[6]}\n                       =  18 + 6\n                       =  24\n\n----------------------------------------------------------\n\nI hope you post more questions like these. They evoke a lot of thought. I hope the solution serves th purpose.
752879
ok re arranged the second question so that the 6 is factored out so its x= such and such then sub the x equals such and such into the first equation and solve then do the same thing to Y in the second equation
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)
1135408
OH NOES THIS IS LIKE A MATH PROBLEM!!!! i can't handle any more variables :( sry.\n\n\nhahahaha.
773695
Look in your text book under "factoring polynomials"... or you could try this web site that explains this fairly thoroughly...\n\nhttp://faculty.ed.umuc.edu/~swalsh/Math%20Articles/Factoring.html
1085664
Good job!  You're halfway there!\n\nAs you are writing it out on your homework, underline the (4x+1) in each of the two terms.  That helps you see what you are factoring.\n\nFactoring out the (4x+1) from each term, you should get your final answer.\n\n=(5x + 4y)(4x+1)
495916
3x^2+9x-84 \nGCF = 3\n3(x^2 + 3x - 28) now we factor the binomial\n\nwe could do this by multiplying the first coefficient (1) by the last coefficient/term (-28). we get -28\nnow we need to select two numbers that multipli to -28 and add to the middle coefficient, 3.\n\nthe two numbers are (7) and (-4)\nwe rewrite the problem with these two terms\n(x^2 + 7x - 4x - 28) i did not write the GCF 3 just yet\nfactor by grouping\n\nx(x + 7) - 4(x + 7)\n\nfactor completely\n\n(x + 7)(x - 4)\n\nfinal answer, don't forget about the 3\n\n3(x + 7)(x - 4)\n\n----------------------------\n4x^2 - 4x - 3   \n1. check for GCF - none\n2. first and last (4)(-3) = -12\n\ntwo numbers that multiply to -12 and add to -4 = 2 & -6\n\n4x^2 + 2x - 6x - 3\n\nfactor by grouping\n\n2x(2x + 1) - 3(2x + 1)\n\n=(2x + 1)(2x - 3)\n\n----------------------------\n3x^2-17x+22 \n\n1. check for GCF - none\n2. first and last   (3)(22) = 66\n3. two numbers are (-6) & (-11)\n4. rewrite\n\n3x^2 - 6x - 11x + 22\n\n5. factor by grouping\n3x(x - 2) - 11(x - 2)\n\n(x - 2)(3x - 11)\n\n----------------------------\nx^2-9 \n\ndifference of squares\n\na^2 - b^2 = (a + b)(a - b)\n\n1. GCF - none\n2. factor\nx^2-9 = (x)^2 - (3)^2 = (x + 3)(x - 3)\n\n---------------------\n\n-3x^2+48?\n\nalso difference of squares\n1. GCF = -3\n\n= -3(x^2 - 16)  \n\n2. factor: x^2 - 16 \nx^2 - 16 = (x)^2 - (4)^2 = (x + 4)(x - 4)\n\n3. final answer = -3(x + 4)(x - 4)
781910
not possible to factor easily ... try the quadratic formula.
858963
The answer is actually much more complex than one would think! The correct answer though is ...\n\n((2^n)(n-.5)!)/(2(.5!))\n\nThat looks kind of messy, so in English that is "2 to the n times n minus 1 half factorial over 2 times 1 half factorial."\n\nPlugging in 1,3,5,7 for n on a graphing calculator, we get 1,3,15,105, which are the correct first four terms in the sequence.
158047
Pretty slick, RED\n\nYou came on pretending that "someone else" had a problem and even gave out a wrong answer on purpose.  Clever!\n\nI can see through that, you just wanted people to give you the right answer without giving you the whole "Do your own homework" speech.\n\nNice! Score one for you.  Unless, of course, I am giving waaaay too much credit for such an ingenious plan.  In which case, x=2, not 24.
892068
Well, you got me stumped!\n\n-------------------------------------\n\nI started working it out and got pretty far.\n\nThe above given information might be incorrect.  Maybe?  I don't know.\n\nI tried it several times.  I even double and triple checked my work!\n\n---------------------------------------------\n\nThis is as far as I got and then I got stumped:\n(D-E) (B+C) = (10J + AFGH) / A\n\n------------------------------------------------------\n\nGreat question!\n\nI look forward to the continuing discussion.\n\nI love the way you worded your question.  Right on!!!\n\nKeep those challenge questions coming!  ;o}
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