Find the partial fraction decomposition of: x^3+x^2+2 divided by (x^2 +2)^2?
600301
x^3+x^2+2 = x^5+2\n\n(x^2+2)^2 = x^4+4\n\n(x^3+x^2+2)/[(x^2+2)^2] = (x^5+2)/(x^4+4) = (x+2)/4 = (x+1)/2
798648
(x+3)(x^2+2x+4)\n\nStart by taking the first term in the first parenthesis and multiplying it by each term in the second parenthesis:\n\nx*x^2 + x*2x + x*4 = x^3 + 2x^2 + 4x\n\nNow take the second term and multiply it by each term in the second parenthesis:\n\n3*x^2 + 3*2x + 3*4 = 3x^2 + 6x +12\n\nNow add all of that together\n\nx^3 + 2x^2 + 4x + 3x^2 + 6x +12\n\nand combine terms\n\nx^3 + 5x^2 + 10x +12
743452
Assuming that x^4 is not in the denominator:\n\n(x-2/x^2) simplifies to  (x^-1) - (2x^-2)\n\nTherefore, the problem becomes:\n\nint[ x^-1  -  2x^-2  + x^4 ]\n\nNow you can use the power rule to integrate each term:\n\nln(x) + 2x^-1 + (x^5)/5  +  C
1054627
assuming this is (3a+1)/(a^2-1) - 1/(a+1)\n\nfactorise the (a^2-1) to give (a+1)(a-1)\n\nmultiply the second fraction by (a-1)\n\n-->  (3a+1)/[(a+1)(a-1)] - 1(a-1)/[(a+1)(a-1)]\n\nfactorising this gives [(3a+1)-(a-1)]/[(a+1)(a-1)]\n\nwhich equals (2a+2)/[(a+1)(a-1)]\n\n= [2(a+1)]/[(a+1)(a-1)] \n\nthe (a+1)'s cancel out...\n\ntherefore = 2/(a-1)
744506
You have \nx^3 = x^2\nSubtract x^2 from each side\nx^3 - x^2 = 0\nFactor out x^2\nx^2 (x-1) = 0\nx^2 = 0 or x-1 = 0\nx = 0 or x = 1\n\nYou do not want to divide both sides by x^2 because then you will lose a solution.  Suppose I tried solving it this way:\nx^3 = x^2\nDivide both sides by x^2\nx^3/x^2 = x^2/x^2\nx = 1\nI only have one solution if I solve it this way.  The solution x=0 was lost when I divided by x^2\n\nNote that x^2 divided by x^3 is 1/x
853619
There is a theorem related to polynomial divisions called the remainder theorem. Let p(x) be a polynomial of any degree. When it is divided by a linear polynomial of the form x - a, the remainder is p(a).\nAnswer 1:\nLet p(x) = x^5 - 3x^3 + x^2 - 3\nDivisor = x - 2\nHere, a = 2\np(2) = 2^5 - 3(2)^3 + 2^2 - 3\n= 32 - 24 + 4 - 3\n= 32 - 24 + 1\n= 8 + 1   \n= 9\n\nAnswer 2:\np(x) = x^5 - 3x^3 + x^2 - 3\nDivisor = x + 1\na = -1\np(-1) = (-1)^5 -3(-1)^3 + (-1)^2 - 3\n= -1 + 3 + 1 - 3\n= 0\n\nYou can check with actual division if you like. Looks like some of us have it wrong.
258044
divide the exponent by two
893371
Factor the denominators\nx^2 + 3x + 2 = (x+2)(x+1)\nx^2 + x - 2 = (x+2)(x-1)\n\nThe common denominator must have each factor that occurs in either denominator, so it is (x+2)(x+1)(x-1)\n\nThe first fraction is missing the (x-1) so multiply top and bottom by (x-1).  Do not multiply out the bottom, just the top:\n(x+5)(x-1) = x^2 + 4x - 5\n\nThe second fraction is missing the (x+1) so multiply top and bottom by (x+1) (again just do out the top)\n(x-7)(x+1) = x^2 - 6x - 7\n\nNow since both fractions have the same denominator, add the like terms on top.  This will be 2x^2 - 2x - 12\n\nThis simplifies to 2(x^2 - x - 6) which is 2(x-3)(x+2)\n\nThe (x+2) on top cancels the (x+2) on the bottom.\n\nFinal answer: [2(x-3)] / [(x+1)(x-1)]\n\nThe restrictions are any values for x that cause the denominator to be zero.  This is before reducing!  So x cannot be 1, -1, or -2.
904672
let u = x^2 + 1, then you have du = 2xdx, => dx = 1/2x du,\n\nsubstitute back you are on the way  to solve it
685531
[(2x^3+2x^2+3x+3)/(x^2-9)]*\n[(x^2+2x-15)/(4x^2+6)]\n\nFactor each expression individually.\n\n(2x^3+2x^2+3x+3) = (x+1)(2x^2+3)\n\n(x^2-9) = (x+3)(x-3)\n\n(x^2+2x-15) = (x+5)(x-3)\n\n(4x^2+6) = 2(2x^2+3)\n\nSubstituting these in, our expression becomes\n{[(x+1)(2x^2+3)]/[(x+3)(x-3)]}\n*{[(x+5)(x-3)]/[2(2x^2+3)]}\n\nCombining these into one larger expression, we get\n[(x+1)(2x^2+3)(x+5)(x-3)]/\n[2(x+3)(x-3)(2x^2+3)]\n\nCancel the quantities common to the numerator and denominator and we get our final answer\n[(x+1)(x+5)]/[2(x+3)]\n\nGood luck.
842257
divide through\n\ny = [x^(3/2)]/3 - 2x^(1/2) -x^(-1/2)\n\nthen take the derivative\n\ndy/dx = [x^(1/2)]/2 - x^(-1/2) + [x^(-3/2)]/2\n\nthen put x^(3/2) in the denominator\n\ndy/dx = [1/x^(3/2)]*{ [x^2]/2 - x + 1/2 }\n\nwe take out the one half to find\n\ndy/dx = [1/2x^(3/2)] * [ x^2 - 2x + 1 ]\n\nwhich is the same as writing \n\ndy/dx = [1/2x^(3/2)] * (x - 1)^2\n\nor a = -1 and b = 2
28370
x(x-3)/x(x+2) divided by (x-3)(x-2)/(x-2)(x+2)\nfrom first fraction curtail x \nfrom second fraction curtail (x-2) and we have\n(x-3)/(x+2) divided by (x-3)/(x+2), which is 1\nx^2 - 5x + 6 = (x-2)(x-3)\nx^2 - 4 = (x -2)(x+2)
697928
First you group the 'like terms'; in your example, the first two and the last two. Then you see the common factor and pull it out\n\ny^2(x^2-1) - 15 (x^2-1)\nNow you see the common term in both and pull that out\n(x^2-1) (y^2-15)\n\nthus you have your two factors of the polynomial
599364
x^2 + x = x(x+1)\n\nIt can be simplified to x(x+1).
443446
S (x^2-4x) / (x^3-6x^2) dx\n\nlet u = x^3-6x^2\nthen du = 3x^2-12x\n\nS (x^2-4x) / (x^3-6x^2) dx = (1/3) du/u\n= (1/3) ln u + C\n= (1/3) ln (x^3-6x^2) + C\n\nAnswer B.
562954
You simply divide each term in the polynomial by that integer.. For your example here it goes..\n\n(3x^2+8x-2) divided by 2 \n\n> {3(x^2)/2} + {8x/2} - {2/2}\n\n>> Answer: 1.5 x^2 + 4x -1    or   (3/2)x^2 + 4x -1
858168
1. 14x^2 +21x\n\na good way to start would be to factor out any common terms:\n\n7(2x^2 + 3x)\n\n7x(2x + 3)