The answer is given but we have to show how to arive with the answer using long division and well so far I can't figure out the step they did to get that answer.  Help would be greatly appreciated.
198872
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).
99937
although the equation is not complete and therefore the full answer can not be given. i believe this equation will need to be differentiated using the product of a product rule.
459997
Have you tried these?  Surely you've been told that division by 0 isn't possible, so there's your answer for 7.  I'm not doing any others on Yahoo answers (too many at once), but if you really need help and are willing to do as much as possible for yourself I'll help you if you email h_chalker@yahoo.com.au, but nasty Maths teachers like me won't just hand you 7 answers on a plate!
453548
I understand your frustration in receiving no Answer yet to this question, but since I do not have any direct knowledge of how to perform the calculation(s) you need ... I hope the link below serves you well.  It appears to explain the mathematics behind all the concepts you mention.
289766
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.
288616
It's hard to format using this system.  It's pretty much just like long division with regular numbers.  First you need to rearrange them so the coefficients of x are decreasing.  Also, in the numerator, fill in any missing powers of x with a 0.\n\n5x^4 + 0x^3 + 10x^2 + 0x + 5\n\nNow you will divide by x-5\n\nx-5| 5x^4 + 0x^3 + 10x^2 + 0x + 5\n\nJust ignore the -5 and look at the x.  One x goes into 5x^4 how many times?  ***5x^3***.  So that is the first part that goes on top of the long division sign.  Now, same as with numbers, multiply this by the x-5 and put it underneath and subtract: \n5x^3 * (x-5) = 5x^4 - 25x^3.  So subtracting this from 5x^4 + 0x^3 + 10x^2 + 0x + 5 you get  25x^3 + 10x^2 + 0x + 5. (Note the 25x^3 is positive because we subtracted a negative number).\n\nNow again, x goes into 25x^3 how many times?  ***25x^2***.  Multiply 25x^2 * (x-5) = 25x^3 - 125x^2.  Subtract and you are left with 135x^2 + 0x + 5.\n\nx goes into 135x^2 ***135x*** times.  135x * (x-5) = 135x^2 - 675x.  Subtract and get 675x + 5.\n\nx goes into 675x ***675*** times.  675 * (x-5) = 675x - 3375.  Subtract: 3380. This is the remainder, since x does not go into 3380 at all, and so it goes over x-5 (just like with plain numbers).\n\nSo then the final answer is to gather up all the numbers across the top of the long division bar, and add on the remainder:\n\n5x^3 + 25x^2 + 135x + 675 + 3380/(x-5)
188908
(3x^3-7x^2+x-20)/(x-3) = 3x^2+2x+7 with a remainder of 1\n         \n         3x^2+2x+7 \n     _______________\nx-3 | 3x^3-7x^2+ x-20\n      -3x^3+9x^2\n    __________\n          2x^2+ x\n         -2x^2+6x\n         ________\n              7x - 20\n             -7x + 21\n             ________\n                   1\n\nI tried to show what this would look like as a division problem but it keeps jumbling up.
551385
Gopal has the correct answer! \n\nSo... becareful with some other results... and... never use:\n0.833 instead of 5/6... it's a mistake!\n\nFinished Algebra with 19!
273732
Division by zero is undefined, even if you divide zero into zero.\n\nhttp://en.wikipedia.org/wiki/Division_by_zero
434430
You typed exponents just fine!\n\nx^3 - x^2 -x + 7 divided by x+1 is a "long division" type problem rather than factoring because x+1 doesn't go evenly into the polynomial\n\nThe answer is x^2 -2x + 1 + 6/x+1
664983
the 1/3 will be negative. (if you are referring to the "please help again" question)\n\nAll you have to do is solve the equation as if there is an = sign in the middle. keep track of how many times a negative number was multiplied or divided, then switch the sign in the middle that many times
45011
Use a calculator - numerator divided by denominator will give you the answer.
435310
I prefer "long" multiplication but it doesn't line up as well...\n\n3x^2 + 7x - 6\n* _____x+12\n--------------------  Multiplying term by term\n_____36x^2 + 84X - 72\n3x^3 + 7x^2 - 6x\n--------------------------------  Add term by term\n3x^3 + 43x^2 + 78x - 72   <--  final answer
611709
It *is* awfully hard to write algebra expressions on Yahoo Answers, isn't it? Oh well, we do what we can.\n\nAddressing your first question, let's write it out like a long division problem.  You'll have to imagine the horizontal bar above the larger polynomial, because I can't format it properly here:\n\nx^2 + 3x - 2  )  x^4 + x^3 - 5x^2 + 13x - 6\n\nOur first step is to look at the first term in each polynomial, and divide the smaller into the larger.  x^4 / x^2 = x^2, so the first term we write in our quotient (above the bar) is x^2.  You can write it directly above the x^4, if you wish, or above the 5x^2 term, your preference.\n\nNow, we're going to multiply that "partial quotient" by the entire divisor (x^2 + 3x - 2), which gives us x^4 + 3x^3 - 2x^2.  Write that "partial product" below the first three terms of the original dividend (x^4 + x^3 - 5x^2).  Our next step is to subtract the partial product: (x^4 + x^3 - 5x^2) - (x^4 + 3x^3 - 2x^2) = -2x^3 - 3x^2.  We then drop down the next term from the dividend, the 13x term, and repeat:\n\n-2x^3 / x^2 = -2x (add that term to our quotient)\n\n-2x * (x^2 + 3x - 2) = -2x^3 - 6x^2 + 4x\n\n(-2x^3 - 3x^2 + 13x) - (-2x^3 - 6x^2 + 4x) = 3x^2 + 9x\n\nFinally, we drop down the last term from the dividend, -6, and repeat once more.  I'm sorry that I can't write this out the way your math book can... I really hope Yahoo Answers creates some sort of way of inputting formulas, or at least monospacing text entry!\n\n3x^2 / x^2 = 3 (so we add a +3 term to our quotient)\n\n3 * (x^2 + 3x - 2) = 3x^2 + 9x - 6\n\n(3x^2 + 9x - 6) - (3x^2 + 9x - 6) = 0... no remainder!\n\nSo our final quotient is x^2 - 2x + 3, with no remainder.\n===\nNow, when doing the 2nd problem, you'll want to write the larger polynomial, the dividend, as x^4 - 3x^3 + 0x^2 + 6x - 5.  We incorporate that extra 0x^2 term so that we never find ourselves trying to subtract unlike terms.\n\nSorry there's no better way to write this out on Yahoo Answers.  Hope that helped anyway!
426350
The answer is i.\n\nFirst, look at i^0(i to the 0th power). It is 1, because anything to the 0th power is 1.\nNow, look at i^1(i to the 1st power). It is simply i.\nNext, look at i^2, which is -1, by the definition of i.\ni^3 = i^2 * i, right, which is -1 * i = -i.\ni^4 = 1 etc..\n\nObserve i^n, where n is any value over 3. We can then write i^n as i^4 * i^(n-4) = 1*i^(n-4) = i^(n-4). So i^n = i^(n-4). So we can just keep subtracting 4 from n until we are at 0,1,2 or 3, which we now know! We see that this number (0,1,2 or 3) is simply the remainder when we divide n by 4.\n\nSo to find powers of i, we simply find the remainder when it is divided by 4, and pit it against this chart we have made:\n\n0:1\n1: i\n2: -1\n3: -i\n\nSo for this problem, if we do the long division, we see that 21/4 = 5 + 1/4, so the remainder is 1. Therefore, the answer to this particular problem is i.\n\nSo the answer is i.
643683
this question will require calcs using logarithm.\nhence will take time
105050
Have patience :)  You got some answers the first time you asked this.  Here it is again:\n\nIt does not get eliminated, it gets split up into the 2 factors.\n\nOn this one, you can go ahead and divide by 2 to make things a little easier. You'll be left with\n\n3x^2 - 7x - 6 = 0\n\nNow to factor. You need 2 numbers whose product equals 3*-6, or -18 and whose sum equals -7. From this point on it's just sort of educated guessing. \n\n1*-18 = -18, but 1-18 = -17, so that won't work\n2*-9 = -18, and 1-9 = -7 -- THIS ONE WORKS!! yay!\n\nSo now you can rewrite your problem as\n\n3x^2 - 9x + 2x -6 = 0 and group the first 2 and last 2 terms\n\n(3x^2-9x) + (2x-6) = 0 -- you can factor a 3x out of the first group, and a 2 out of the second\n\n3x(x-3) + 2(x-3)=0 -- and now you can factor an (x-3) out\n\n(x-3)(3x+2)=0\n\n~~~~~~~~~~~~~~~~~~~\nOops, you're right. I was typing too fast.  Oh well, I still got the right final answer :)  Thanks