Gah! Factoring!!! (I'm a junior and never learned this. It has caught up wiuth me now.)?
98662
purplemath.com is good so is sparknotes.com
768295
Using an equation is the lazy, mechanical way to do it. If you want to understand what's going on, factor it if possible.\n\nIf you *really* want to know what's going on, derive the quadratic formula yourself. It isn't as hard as it looks, and it can be done on one page.
221530
Add them together and divide by 5.\n\nWow. I could seriously have you people do my homework for me. Every night. I'll type out the problems and see how you all do. :P JP
559559
Multiplication and division are basically just a matter of memorization.  I know if you're just learning it, its hard.  I remember having trouble as well when I first started.  But keep at it.  It will get frustrating at times, but keep at it.  The problem with math is that as you go farther, you need to remember everything you've already learned.  But otherwise, it can be real easy.  Just keep at it and you'll do fine.
815517
OK First you take out the common factor which is y. After you have done that.you have (y^2-12y+36). After this, it becomes simple factoring. The answer is\ny(y-6)(y-6)\nor also known as y(y-6)^2
607601
I think you set the equation to zero then you have to factor it.  Go to the link below if you wanna know how to do it. OR look your your math book.
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
729558
I worked this problem. I did get both sides to zero. I don't even know how to begin explaining it.\n\nFor the sake of typing this with some readability, let (k+1) = b and (k+2) = c.\n\nFirst you subtract (2^c)/(k+3) from both sides to get 0 on one side. Then to get a common denominator, you multiply top and bottom of the first term by (k+3) and top and bottom of the third term by (k+2). You get:\n\n[ (k+3)(2^b) + (k+1)(2^b) - (k+2)(2^c) ] / [ (k+2)(k+3) ] = 0\n\nThen you distribute out the terms:\n\n[ k(2^b) + 3(2^b) + k(2^b) + (2^b) - k(2^c) - 2(2^c) ] / [ (k+2)(k+3) ]\n\nThen you factor out 2^b from the first four terms:\n\n[ (2^b)(k + 3 + k + 1) - k(2^c) - 2(2^c) ] / [ (k+2)(k+3) ]\n\nThen you combine like terms on the left four. On the right two, you reduce the exponent by one and multiply the term by two. Since c is one more than b, this turns (2^c) into (2^b). You can do this since it's an exponent on two:\n\n[ (2k + 4)(2^b) - 2k(2^b) - 4(2^b) ] / [ (k+2)(k+3) ]\n\nYou can factor 2^b out of the right two terms to get:\n\n[ (2k + 4)(2^b) - (2k + 4)(2^b) ] / [ (k+2)(k+3) ] = 0\n\nThe two terms in the numerator cancel out:\n\n0 / [ (k+2)(k+3) ] = 0\n\nAnd:\n\n0 = 0\n\nI must admit that I liked this problem. I don't do math for fun, but every once in a while I like a challenge.\n\nI hope this helped.
745882
C is correct but not complete... only the 3 is factored out... the real answer would be:\n\n3(x-1)(x+4)
561262
u jammed the equation together.\ndo u mean\n6x^3 + ( 6/x - 2 * x^2 - x/3x2 - 3x + 3)\nits really hard to see ur question. can u write it out with spaces...\nhurry up i am waiting... lol
610613
In 6ab + 4a + 3b + 2, we group the first 2 terms and the last 2 terms:\n\n6ab + 4a + 3b + 2 =\n(6ab + 4a) + (3b + 2)\nthen factor out the common factor 2a in (6ab + 4a)\n(Note: to factor out is the same as dividing each term with the common factor)\n\nwe get:\n2a(3b + 2) + (3b + 2)\nNotice that in each term of the above we have a common factor of (3b + 2)\nBy dividing each term by (3b + 2) or factoring out (3b + 2) we have:\n(3b + 2)(2a + 1)
179350
ok well u gotta do long division for this. but first you need to find a factor. theres some theorem or rule or somethin that sayz, basically check all the numbers that are factors of the last number. so the last number is -12. factors are +-1, +-2, +-3, +-4, +-6. so put each one into the equation for x  and eventually, 1 of them will make the equation =0. so i think 1 works. so that means the factor is x-1. now u do long division.\n\nso first you ignore the -1 and u just see what do u multiply x by to get x^4. and its x^3. and then you multiply the whole   (x-1) by x^3   and subtract it from your equation. and then bring down the next term. and repeat it and you should not get a remainder.\n\n....................x^3 - 2x^2 - 11x + 12\n(x-1)...%...(x^4 - 3x^3 - 9x^2 + 23x - 12)\n.............x^4 - x^3\n\n...............0 -2x^3 - 9x^2\n.................-2x^3 + 2x^2\n\n...................0   -11x^2 + 23x\n.......................-11x^2 + 11x\n\n........................0     + 12x - 12\n.................................0 + 0\n\nand so there. your first factor is (x-1) and now u gotta repeat the same thing for your quotient. which is  \nx^3 - 2x^2 - 11x + 12.  so 1 works here again. which basically means (x-1) factors out twice.    so divide again\n\n....................x^2 - x - 12\n(x-1)...%...(x^3 - 2x^2 - 11x + 12)\n.............x^3 - x^2\n\n..................-x^2 - 11x\n..................-x^2 + x\n\n........................-12x + 12\n........................-12x + 12\n\nand now u have a quadratic equation x^2 - x - 12. which isnt that hard to solve. 2 numbers that add to -1 and multiply to -12  is just -4 and 3. so (x-4)(x+3)\n\nso the whole equation = (x-1)^2 (x-4)(x+3)    so the   zeros   are  1, 4, -3\n\nhope u get it. let meknow if u need more help. laterz
860900
theres a lot of things. if you have x amount of money in the bank and need y amount of money for rent and food then spending z amount of money on clothes will keep you in your spending limit! haha! or if for some reason your boss decides to build a wall and you have x & z but the equasion is x + y= z. i dunno man. algebra sucks but until they take it off the needed for graduation list your stuck with it.
208764
This is actually elementary math. If you remember the order of operation, it helps. \n\nStep 1: Do the part of the problem in the parenthesis.\n  Example: (4/3)= 1.3333333333 (you get where this is going)\n\nStep 2: Take that answer and multiply it by the named multiplier.\n  Example: 1.33333 * 12= 16\n\nStep 3: Take the result of that transaction and devide it by the named devisor. For this method don't use the fraction way, and don't use a calculator. Do a basic devision problem like you were taught in the 3rd grade using the "bracket."\n  Example: 16/13 = 1  (13 goes into 16 1 time)\n\nStep 4: Subtract the 13 from the 16 and it will give you your remainder.\n  Example: 16/13 = 1   16-13= 3\n\nThe answer when written correctly isn't with a decimal but would look like this- 1 R3.\n\nAlso for a little note for this- Order of operation:\nP (parenthesis)\nE (exponents)\nM (multiply)\nD (devide)\nA (addition)\nS (subtraction)\n\nHope this helps!
781910
not possible to factor easily ... try the quadratic formula.
835122
This sounds like a homework question and I think you need to figure it out yourself -- this is how you learn!  What I can tell you as a hint is to start at the beginning and then add or subtract the numbers to change the balance, and break it down so you do it is step by step.
609611
Dear amit.wizkid, thou little knoweth what an abysmal error thou maketh.\n\ntut_einstein, you are better off noting that the sum of the first 'n' terms of an AP, whose first term is 'a' and the difference between terms is 'd' is :\n\nS = (n/2)[2a + (n-1)d]\n\nYou state that \nS1/S2 = (n+1)/(n+3)\n\nwhich is \n\n[2a1 + (n-1)d1] :[2a2 + (n-1)d2] = (n+1) : (n+3) ..... (I)\n\nRemember that what you desire to know is\n(a1 + 5d1) : (a2 + 5d2) \n\nPut n = 11 in (I)\n\n[2a1 + 10 d1] :[2a2 + 10 d2] = 12:14\n\nwhich gives\n\n[a1 + 5d1] : [a2 + 5d2] = 6:7\n\nwhich is what you desire. \n\nShould suffice.