Find the quotient and remainder using long division?
913334
x^2 + 2x + 7 Remainder 10\n\ni can add the steps if you don't think i did it by hand
762587
use remainder theorem\nhttp://en.wikipedia.org/wiki/Remainder_theorem\n"When p(x) is divided by (x-a), the remainder is p(a)"\nUR QUESTION:\np(x) = x^3+3x+k\n(x-a) = (x-1)\np(a) = 6\nhence, p(1)=6\nso, (1)^3 + 3(1) + k = 6\nhence, 1+3+k = 6\nso, k = 2\n\nhence, k = 2
275765
you need to know the relation that the greeks found thousands years ago Δ=δπ+υ. this is the identity of the incomplete division Δ is the number you want to divide,\nδ is the divider, π is the quotient and υ is the remainder.\nyou just can´t round up or down the remainder
904822
Hello Dear Blue \n\n2x^3+8x^2 - 5x + 5 ÷ x-2\n\n▪ Step1\n( 2x^2 ) ( x-2 ) = 2x^3 - 4x^2 ; { let Q1 = 2x^2 }\n2x^3+8x^2 - 5x + 5 - ( 2x^3 - 4x^2 ) = \n2x^3+8x^2 - 5x + 5 - 2x^3 + 4x^2 = +12x^2 - 5x + 5 {remainder} \n\n▪ Step 2\n+12x^2 - 5x + 5 ÷ x - 2 \n(x - 2) (12x) = 12x^2 - 24x ; { let Q2 = +12x }\n+12x^2 - 5x + 5 - ( 12x^2 - 24x  ) =\n+12x^2 - 5x + 5 - 12x^2 + 24x = +19x +5 {remainder}\n\n▪ Step 3 \n19x +5 ÷ x -2\n(x -2) (+19) = +19x - 38; { let Q3 = +19}\n19x +5  - (+19x - 38) =\n19x +5  -19x + 38 = +43 {remainder}\n\n▪ Step 4\nSum Q1, Q2, Q3\nQ1 + Q2 + Q3\n2x^2 +12x +19\n\nSo;\n(2x^3+8x^2-5x+5) ÷ (x - 2)    = >\n\nQ(x) OR quotient = 2x^2 +12x +19 \nR(x) OR Remainder is + 43 \n\nGood Luck dear.
742127
= (x^12)/12 - (8x^11)/11 + (5x^10)/2 - (10x^9)/3 - (15x^8)/4 + 24x^7 - 49x^6 + 60x^5 - (195x^4)/4 + (80x^3)/3 - (19x^2)/2 + 2x + C
823706
2x^2 + 13x + 39 remainder 122\n\ndoes no one know what synthetic division is?  i suppose i only learned it last year...
538734
It's just like long division... but it is hard to show in text form here.\n\nLet's start with the second one, since it is simpler:\n\n2x^2 + 6x - 5 / 2x - 4\n\nWrite it as you would for long division:\n\n```````` ____________\n2x - 4 ) 2x^2 + 6x - 5\n\nStart with the first couple terms.  How many times does 2x-4 go into 2x^2 + 6x.  It goes in once (times x), with a remainder of 2.\n\n```````` _______1x____\n2x - 4 ) 2x^2 + 6x - 5\n`````````-` 2x^2 - 4x\n`````````----------------------\n````````````````````10x - 5\n\nAnd how many times does 2x-4 go into 10x-5... 5 times.\n\n```````` _______1x_+5_\n2x - 4 ) 2x^2 + 6x - 5\n`````````-` 2x^2 - 4x\n`````````----------------------\n````````````````````10x - 5\n``````````````````-`10x -20\n`````````----------------------\n```````````````````````````15\n\nSo the answer is x + 5 with a remainder of 15/(2x-4)\n\nTry the first on your own... it doesn't even have a remainder...\n\nHere's some hints:\n\nx + 11 ) 3 x^3 + 35 x^2 + 19 x - 33\n\nIgnoring x for a moment\n\n1 11 ) 3 35 19 -33\n\nFirst one goes in 3 times\n\n1 11 ) 3 35 19 -33\n```````- 3 33\n```````-----------------\n`````````````2 19\n\nNext goes in 2 times\n\n1 11 ) 3 35 19 -33\n```````- 3 33\n```````-----------------\n`````````````2 19\n```````````- 2 22\n```````-----------------\n```````````````-3 -33\n\nThis one goes in -3 times\n\n1 11 ) 3 35 19 -33\n```````- 3 33\n```````-----------------\n`````````````2 19\n```````````- 2 22\n```````-----------------\n```````````````-3 -33\n`````````````- -3 -33\n```````-----------------\n````````````````````0\n\nPut x back in you have 3x^2 + 2x - 3
984538
(2/6) / (4/9)\n1/3 * 9/4\n3/4
762985
(b-4)/(b-4)(b+4) simplifies to 1/(b+4)\nrestrict denominator...bcannot equal 4 or -4
336691
Wow this is a long question.... hehe.\n\n1) Firstly consider any integer divided by 6, we must get one of the residues ( remainders ), r = 0,1,2,3,4,5. If the remainder of y divided by 6 is r, thus we can write \n y = 6 x "quotient" + "remainder" = 6k + r, where let k denote the value of the quotient. Since r = 0,1,2,3,4,5, we have just shown that any integer y ( we may assume positive ), can be written in one of the "forms"\n y = 6k , 6k + 1, 6k + 2, 6k + 3, 6k + 4, 6k + 5.\n\nNow among these, the ones that are composite ( NOT prime ) \nfor any integer k, are\n      6k , Even\n    6k + 2, Even\n    6k + 3 = 3(2k + 1) , a multiple of 3. \n    6k + 4, Even\n\nRemembering that the only even prime number is 2.\nThus, the only possibilities for forms for primes in here is either 6k+1 or 6k + 5 . All other forms denotes a composite integer. Remember that a composite integer is an integer with at least two distinct prime factors.\n\nIf p is form 6k+1, Let p = 6k+1,\nThus, p^2 + 2 = ( 6k + 1 )^2 + 2\n              = 36k^2 + 12k + 3\n              = 3 ( 12k^2 + 4k + 1 )\n              = A multiple of 3 > 3 \n              = A Composite number.  \n\nIf p is form 6k+5, Let p = 6k+5,\nThus, p^2 + 2 = ( 6k + 5 )^2 + 2\n              = 36k^2 + 60k + 27\n              = 3 ( 12k^2 + 20k + 9 )\n              = A multiple of 3 > 3 \n              = A Composite number.\n\nThus, in both cases, if p is prime then p^2 + 2 is always composite.\n\n2)\nAssume that the number of primes are finite.\nNow the integer N is NOT divisible by any of the primes in the list p_1, p_2, ..., p_n.\nWhy? because say p_k divides N. Then, p_k must divide the following integer : p_1*...*p_n where it does NOT contain p_k. Which is a contradiction .\nThus, primes  must be infinite for if not, they produce the contradiction shown above.\n\n3)\n\nI will not use bertrand's conjecture because its NOT proven yet... So, i'll skip this one... its redundant. I will not  use something that has not been proven yet... i.e, a conjecture.
1135639
(1 - a/b) divided by (a^2/b^2 - 1)\nGet a common denominator for 1 - a/b\n1 - a/b\n= b/b - a/b\n= (b-a)/b\nGet a common denominator for a^2/b^2 - 1\na^2/b^2 - 1 \n= a^2/b^2 - b^2/b^2\n= (a^2-b^2)/b^2\n= (a-b)(a+b)/b^2\n\n(b-a)/b divided by (a-b)(a+b)/b^2\n= (b-a)/b times b^2/[(a-b)(a+b)]\nCancel anything on the top with anything on the bottom.  Note that a-b = -(b-a).\nAnswer: -b/(a+b)
971390
{3XY^2M/1}-{2M/[X^-1Y^-1]^2\n+{[4X^2Y^2]/M^-1}-\n(3X^(3Y)^(3M)^-1/(XYM)^-2
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
1166733
dy/dx = {(3x^2-2)(4x-3)^4 - (x^3-2x)(4)(4x-3)^3(4)}/(4x-3)^8\n= {(3x^2-2)(4x-3)^4 - 16(x^3-2x)(4x-3)^3}/(4x-3)^8\n\nUsing the quotient rule.  d/dx f(x)/g(x) = {f'(x)g(x) - g'(x)f(x)}/g(x)^2
256280
divide by -4\n80,-20,5,-5/4,5/16,-5/64....\n{n, n/(-4)}
161655
Divide the decimal number by 16.   Treat the division as an integer division.   \nWrite down the remainder (in hexadecimal). \nDivide the result again by 16.  Treat the division as an integer division.   \nRepeat step 2 and 3 until result is 0. \nThe hex value is the digit sequence of the remainders from the last to first. \ncheck here: http://www.permadi.com/tutorial/numDecToHex/
699646
The easiest way is to find the GCF of the numerator and denominator.And then divide the numeator and denominator by the GCF and write the quotient s as the new numerator and denominator.