Does anyone know about long division out there? Plese I need help!!!?
402291
if you have 5 divided by 235 you would first want to see if 5 could go into the 2 since it can't you would see if it could go into 23 since it can do simple division 5 goes into 23 4 times with 3 left over. you would then divide the 5 by what is left which is the 3 and the 5. taking the 5 it goes into 35  7 times which gives you an answer of 47.  if this doesn't make much sense i have found a website that might help you.
607598
Fractions were always the hardest for me, and I still have to do a quick check! Rather than go through each one (add, subtract, multiply, subtract, I’ll send you a really good web site. http://www.mathleague.com/help/fractions/fractions.htm
811480
1)  I'm afraid I don't understand the question.  It's possible that  you've been taught a different technique than I teach.  The method I teach is essentially the long division you learned in third/fourth grade, modified to work with polynomials.  I'm afraid I'm no help there.\n\n2)  Those are patterns.  You need to understand them and learn to recognize them.  For instance:\n\nx^2 - y^2\n\nis a difference of squares.  That means it will factor into (x+y)(x-y).\n\n9a^2b^4 - 4x^6\n\nis also a difference of squares.  It will factor into\n\n(3ab^2 - 2x^3)(3ab^2 + 2x^3)\n\nThat's what I mean by being able to recognize them.\n\n3)  The only other thing that's important is that the pattern repeats forever.  i^5 = i, i^6 = -1, and so on.  So, i^400 = 1.\n\nAs for the rest of question 3 - well, it depends on what you're being asked.  If you're asked to find all of the rational solutions, then of course you'd find the rational solutions.  If you're asked to find ALL of the solutions, that would certainly include the rational ones.  If you're asked to find all imaginary or complex solutions, then you wouldn't necessarily look for the rational ones, but it might make it easier to solve the problem if you did.  As a thumbrule, I'd suggest always looking for rational solutions unless you have a specific reason not to.\n\nAs for your additional question:  x^2 + 16 = (x+4)^2 just doesn't work.  Try reversing it:\n\n(x + 4)^2 = (x + 4)(x + 4) = x^2 + 4x + 4x + 16 = x^2 + 8x + 16\n\nYou're missing the cross-terms if you try to factor x^2 + 16 that way.  In fact, a sum of squares does not factor over the real numbers.\n\nIf you're ever unsure why a particular factoring pattern doesn't work, you will almost always get an answer by trying it the other way.
93955
change mixed fractions to proper fractions and then divide,\nwhats so hard?
443017
Set it up like a long division problem:\n\nx - 1 ) x^5 + x^4 - 2x^3 + 2x^2 - 3x - 7\n\nNow focus on the x terms.  x is on the outside, and x^5 is on the inside.  Just like long division with numbers, how many times does x go into x^5?  x^4 times.  That goes on top, and then you multiply that by x - 1 and subtract, again like normal long division.\n\nx^5 + x^4 - 2x^3 + 2x^2 - 3x - 7\n-(x^5 - x^4)\n--------------------\n2x^4 - 2x^3 + 2x^2 - 3x - 7\n\n\nRepeat process...x goes into 2x^4 2x^3 times:\n\n2x^4 - 2x^3 + 2x^2 - 3x - 7\n-(2x^4 - 2x^3)\n--------------------\n2x^2 - 3x - 7\n\nx goes into 2x^2 2x times:\n2x^2 - 3x - 7\n-(2x^2 - 2x)\n--------------------\n-x - 7\n\nFinally, x goes into -x -1 time:\n-x - 7\n-(-x + 1)\n--------------------\n-8\n\nSo you have x^4 + 2x^3 + 2x - 1 with a -8 remainder, which you can write as -8/(x - 1).
352026
I would like to be able to give you the answer but I don't think I would be doing you a favour.  If you are to get practice in division, it is better for you to work it out on your own.  Does your teacher know that you are getting homework answers here on Yahoo?
381493
It's set up like a long division problem.  Let's say you have x^2 - 6x - 27 divided by x + 3.  You'd set up like this:\n\nx + 3 ) x^2 - 6x - 27\n\nFor polynomial division, just focus on the variables and the exponents.  To start this one, look at how many times x goes into x^2: x times.  So x would go on top of the dividing symbol, and then you'd multiply the value on top by the value on the outside and subtract, just like a long division problem.  So you'd end up with\n\nx^2 - 6x - 27\n-(x^2 + 3x)\n----------\n-9x - 27\n\nRemember that, just like long division, you're subtracting, so the problems becomes x^2 - 6x - 27 - x^2 - 3x if you wrote it out this way.  The x^2 terms cancel out, and you'd have to bring down the -27.\n\nNow repeat the process - how many times does x go into -9x?  -9 times.  So the -9 goes on top (making your top x - 9) and your problem look like this:\n\nx^2 - 6x - 27\n-(x^2 + 3x)\n----------\n-9x - 27\n-(-9x - 27)\n----------\n0\n\nSince you have no remainder, your answer is x - 9.  And if you're ever unsure, you can always use FOIL to see if you get your original trinomial:\n\n(x + 3)(x - 9) = x^2 - 9x + 3x - 27\n= x^2 - 6x - 27
985294
you just keep on deriving.\n\nI find it easier to split things into exponents.\n\nI am assuming your question goes\n\n(1/(2x))  - 4 = g(x)\nor\n\n(1/2)*x^-1     - 4 = g(x)\n\ndifferentiate once\n\ng'(x) = -(1/2) * x^-2 \ng''(x) = -2 * (-1/2) * x^-3  = x^-3\ng'''(x) = -3 * x^-4  =  -3/(x^-4)  **ans**\n\nPlug 3 into this... bah\n\nFor reference please use parenthesis to make your problem clearer for us. I just solved the whole thing using the wrong g(x).\n\n\n===!using the correct g(x) which should be g(x) =1 / (2x -4)===\n\ng(x) = 1 / (2x -4) = (2x - 4)^-1  don't forget the chain rule\n\ng'(x) = -1*2*(2x - 4)^-2 = -2 * (2x -4)^-2    chain rule again\n\ng''(x) = -2*-2*2*(2x -4)^-3 = 8*(2x -4)^-3    chain rule again\n\ng'''(x) = -3*2*8(2x - 4)^-4 = -48 * (2x -4)^-4\n\ng'''(x) = -48/[(2x -4)^4]\ng'''(3) = -48 * (6 -4)^-4 = -48/ (2^4) = -48/16   = ***real ans***\n\ng'''(3) = -3
210386
I think the point may be to use synthetic division.  Look that up in your text.
978661
Have fun:\nhttp://www.math.nyu.edu/~griffith/teaching/06numanalysis/labs/bisection/
605811
As Lauren said, it's 1/x.\n\nMemorize this. It will be very useful.
967957
There really isn't any way to do this other than guess and check. If you find that you substitute some value in and find f(a) = 0, then (x-a) will be a factor. You can repeat that a few times until you have found some roots, and then do a long division to something which you *can* solve by hand (a quadratic or linear).\nNote also that you only need to check things which are factors of the constant term, and there aren't too many.\nIn the first case, you will find that substituting in -2, -1 and 3 will all give 0, but you may have trouble finding the last root. So you will have to do some long divisions (divide by (x+2)(x+1)(x-3)), and you will end up finding that the -1 was a repeated root - so the answer is (x+2)(x+1)^2(x-3).\n\nSame thing for the second: you will find -1 as one root. You can then do a long division and find -1 is repeated, and again, and again, and again. The actual factorisation is (x+1)^5 - you may have noticed this if you knew much about pascals triangle and recognised the coefficients are an entire row - 1,5,10,10,5,1.
990473
Start with demonstrating the multiplication of fractions through physical models.\n\nSplit a rectangle in half.  That is 1 x 1/2... because you are counting that one thing 1/2 time.\n\nThen split the 1/2 into 1/2.  1/2, counted 1/2 time, is 1/4.  See it?\n\nOK, so... through whole numbers you have already demonstrated (and they should already have shown true comprehension of the idea) that division is the inverse of multiplication. So...\n\n\n1/2 divided by 2/3 is the same as 1/2 x 3/2 because that is the inverse. Division of whole numbers is a reduction of value... division of fractions is an increase of value.\n\nAlso, try this with decimals.\n\n1/10 divided by 4.\n\n1/10 divided by 4/1.\n\n1/10 x 1/4 = 1/40.\n\n1/40 = 0.025.\n\n   ____\n4 |0.1     = 0.025.\n\n25 x 4 = 100\n\n2.5 x 4 = 10.0\n\n.25 x 4 = 1.00\n\n.025 x 4 = 0.100\n\nStart to see the pattern now?
89171
You cant just add or substract whatever is left. You invalidate the function of "division" by partially dividing then filling in the rest of the numerator. The number left over after something has been divided is called the modulo, and is useful for programming functions. You cant do an incomplete division(only dividing 20/5 instead of the 2/5 also) and then speculate as to why it cant be negative when doing a full division. It invalidates the definition.
154380
I posted these sources earlier to someone else who was looking for math help.\n\nGood Luck!!
579225
Division by zero is undefined, even if you divide zero into zero.\n\nhttp://en.wikipedia.org/wiki/Division_by_zero
887079
The long hand method for dividing polonomials is much like normal division\n\n\nput the denominator to the left of your square root looking symbol and then your numerator under your that symbol....\n\n\n1.take the first term from the denominator and muliply it by whatever you need to in order to get the first term in the numerator in this case that is 3x(keep this in some part[if you use normal division method it will be on top of symbol as it is part of your answer)\n\n2. now multiply the operand(3x in this case) by your entire denominator for this= 3x^4 + 3x^3 - 27x\n\n3. now take the numerator and subtract what you got from it\n(3x^4 - 6x^3 - 14x^2 - 6x + 18) - (3x^4 + 3x^3 - 27x) = -9x^3 -14x^2 + 21X + 18\n\n4. now repeat steps 1-3, with your product from #3(every time) as your new numerator, untill you need a negative exponent to complete step 1\n\nyour overall answer is 3x - 9 - {(5x^2 + 21x - 63)/(x^3 + x^2 -9)}\n\nsorry my steps are so confusing but I rarely help others... so I came here to kind of practice for tutoring