5.Find the remainder when 2(x*x*x)-5(x*x)+3x-1 is divided by x+2?
446521
The remainder is -43 because by the remainder theorem, f(c) is the remainder when a polynomial is divided by x-c. In this case, c is -2, so plug it into the equation and the remainder is what you get out of it.\n\n2(-2)(-2)(-2)-5(-2)(-2)+3(-2)-1\n= -16-20-6-1 = -43
1039099
This is the answer i got when i worked out the problem\n1. You have to distribute the (x + 2) & (x + 1)....you do the by using the FOIL method....Firsts Outsides Insides Last and you get: x + 2 = x^2 + 3x + 2\n\n2. Get rid of the 2 on the left side by subtracting it on the right side and you end up with : x = x^2 + 3x\n\n3. Now you can move the x on the left to the right side too so you have: 0 = x^2 + 2x\n\n4. Pull out and x leaving you with 0= x (x+2)\n\n5. x = 0 and x = -2\n\nMy algebra is rusty but I think this is how it's done.
1138309
I hope you are trying to say x+2 = (2/3)(2x-1)..if it is correct then\n\n3(x+2) = 2(2x-1) --> 3x+6 = 4x - 2 --> 6+2 = 4x-3x --> 8 = x
707304
1/(x - 1) = (x + 2)/4\ncross multiply\n4*1= (x-1)*(x+2)\n4 = x^2 +2x - 1x -2\n4 = x^2 + x -2\n0 = x^2 +x -6\n0 = (x+3)(x-2)\n\nso x= either -3 or +2\npositive = +2
714589
x/(x+3) - 2(x-1)/(x+3)(x+2)\n\nx(x+2) - 2 (x-1)\n--------------------\n   (x+3) (x+2)\n\n\nx^2 +2\n----------\n(x+3)(x+2)
40442
(2/X+5)+(3/X+4)=(2x/x^2+9x+20)\n\nX = -.341296
790277
ill give it a shot - im not sure , anyways i got 2 answers..\n\n(1st)\n\n(3-x)x+2 = 2 (x^2-2)\n\n3x-x^2+6-2x = 2x^2-4\n\nx-x^2+6 = 2x^2-4\n\nx-1/x+6 = 2/x-4\n\nx-1+6x = 2-4x\n\nx+4x+6x = 2+1\n\n11x = 3\n\nx = 3/11\n\n\n(2nd possible answer)\n\n\n(3-x)x+2 = 2 (x^2-2)\n\n3x-x^2+6-2x = 2x^2-4\n\nx-x^2+6 = 2x^2-4\n\nx-1/x+6 = 2/x-4\n\nx-1+6/x = 2-4/x\n\n-5 = -2/x\n\n-5x = -2\n\nx = -2+5\n\nx = 3
423594
(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.
831395
Let f(x) = 2x^3 + 7x^2 - 77x - 40 = 0\n\nTry to find factors by remainder theorem\nSo test x = ±1, ±2, ±5, ±8 (factors of - 40) and try to find one factor.\nf(1) = 2 + 7 - 77 - 40 ≠0\nf(-1) = -2 + 7 + 77 - 40 ≠0\nf(2) = 16 + 28 - 154 - 40 ≠0\nf(-2) = -16 + 28 + 154 - 40 ≠0\nf(5) = 250 + 175 - 385 - 40 = 0 so x - 5 is a factor\n\nBy division\nf(x) = 2x^3 + 7x^2 - 77x - 40\n= (x - 5)(2x^2 + 17x + 8)\n= (x - 5)(2x + 1)(x + 8)\n= 0\nSo x = 5, -1/2, - 8
779760
if by this you mean\n\nf(x) = (x^(2/5)) * 5(x - 1), then\n\nf'(x) = (2/5)x^((2/5) - 1)*(5(x - 1)) + (x^(2/5))(5(x - 1)')\nf'(x) = (10(x - 1)/5)x^((2/5) - (5/5)) + 5x^(2/5)\nf'(x) = (2(x - 1))x^(-3/5) + 5x^(2/5)\nf'(x) = ((2x - 2)/(x^(3/5)) + 5x^(2/5)\nf'(x) = ((2x - 2) + 5(x^(2/5) * x^(3/5)))/(x^(3/5))\nf'(x) = (2x - 2 + 5x^((2/5) + (3/5)))/(x^(3/5))\nf'(x) = (2x - 2 + 5x^(5/5))/(x^(3/5))\nf'(x) = (2x - 2 + 5x)/(x^(3/5))\nf'(x) = (7x - 2)/(x^(3/5))\n\n------------------------------------\n\nBut if you mean\n\nf(x) = (x^2)/(5(x - 1))\n\nf'(x) = (x^2)(5(x - 1))^(-1)\nf'(x) = (2x^(2 - 1))((1/5)(x - 1)^(-1)) + (x^2)((-1/5)(x - 1)^(-1 - 1))\nf'(x) = ((2x)/(5(x - 1)) + ((x^2)/(5(x - 1)^2))\n\nANS : f'(x) = ((2x)/(5(x - 1)) - ((x^2)/(5(x - 1)^2)\n\nor you can do it like this\n\nf'(x) = (2x(x - 1) - x^2)/(5(x - 1)^2)\nf'(x) = (2x^2 - 2x - x^2)/(5(x - 1)^2)\nf'(x) = (x^2 - 2x)/(5(x - 1)^2)\n\nor\n\nf'(x) = (x^2 - 2x)/(5(x - 1)(x - 1))\nf'(x) = (x^2 - 2x)/(5(x^2 - 2x + 1))\nf'(x) = (x^2 - 2x)/(5x^2 - 10x + 5)\n\nANS : (x^2 - 2x)/(5x^2 - 10x + 5)
791629
If x = original number\nand\nR1= remainder of x/3\nR2= remainder of x/5\nR3= remainder of x/7\n\nThen  x = (R2 * 21) + (R3 * 15) - (R1 * 35)\n\nSo for your example of 26:\nR1 = 2\nR2 = 1\nR3 = 5\n\nX = (1 * 21) + (5 * 15) - (2 * 35)\n   = 21 + 75 - 70\n   = 26\n\nNotice how the remainder of one number relates to the multiple of the other two. Also, I think you need to start at 7 instead of 3, because there will be fractions and no remainders for the numbers less than 7.\n\nEDIT:\n\nIf you look at the formula again, you will see that it works for your example, also!  =)\n\nBe careful where you plug in your numbers; they aren't in order.
541499
2x(x-5)-3(2x+1)\n2x^2-10x-6x-3\n2x^2-16x-3
321475
*(2x/(x^2-4)) + (5/2-x) - (1/2+x)\n*(2x/(x^2-4)) + (5/2-x)((2+x)/(2+x)) - (1/2+x)((2-x)/(2-x))\n*(2x/(x^2-4)) + (10+5x)/(4-x^2) - (2-x)/(4-x^2)\n*(2x/(x^2-4)) + (8+6x)/(4-x^2)\n\nyou should be able to get it from there
259375
X - 2/3 X =  2X - 5\n\n2X - X + 2/3 X =5\n\n5/3 X = 5\n\nX = 3\n\n\n3 exceeds two thirds of 3 (=2) by 1 which is equal to 5 less twice of three (6-5=1)
652975
(x+4)/[(3x-1)(3x+1)]-\n(x-2)/(3x+1)=\n\n[(x+4)-(3x-1)(x-2)]/\n[(3x-1)(3x+1)]=\n\n[x+4-3x^2+6x+x-2]/\n[9x^2-1]=\n\n[-3x^2+8x+2]/[9x^2-1]
537832
(x/x-2)-(x+1/x)=8/(x^2-2x)\n[x(x)-(x-2)(x+1)]/[x(x-2)]=8/[x(x-2)]\nx^2-(x^2-2x+x-2)=8\nx^2-x^2+x+2=8\nx=8-2=6
670503
The answer is x2+2x-4 with a remainder of 11/(x+1)