Use synthetic division to find quotient and remainder?
491455
(x^2+4x+1)/(x-2) = x+2+3/(x-2)\n\nx  : x^2+4x+1-x^2-2x =2x+1\n2  : 2x+1 -2x-4 =3
1079776
you can simply the first one to (x-a)(x-a) then you can divide that by (x-a) and it equals (x-a)\n\nDo you know synthetic division; cuz you can divide the second problem by (x+2)^2 if you mulitply (x+2)^2 out.
911395
The easy way is to reverse the divisor. \nie 2/3 divided by 5/9 is the same as 2/3 times 9/5 which is \n(2 x 9)/(3 x 5) = 18/15 = 1 & 1/5
1116253
just divide both sides by 0.015...\n\nand you will get...\n\n566.66 \n\nwith a repeating six.
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.
889816
1/rt3 *1/rt2-1/rt3\n[1/rt3](1/rt2-1)\n[1/rt3](1-rt2/rt2)\nrationalising\n[(1-rt2)*rt6]/6
526107
You are only allowed to factor if the equation is equal to zero; your's is equal to 4.  You need to solve "x^3 - 6x^2 +9x - 4 = 0"  using synthetic division.
163804
Continuously divide the decimal number by 16 and stack up the remainders at each stage, until you finally reach a divisor that is less than 16. Of course, you have to write the remainders above 9 in the alphabetical form - 10=A, 11=B and so on till 15=F. Finally, string the final divisor followed by the stack of remainders read backwards to obtain your hex number.\n\nAlternately, if you are good at converting between decimal and binary, do that and then read off the binary bits from the end in groups of four in the equivalent hex. For example, if after decimal-to-binary conversion, you were to get\n11100011010101, then group them as\n(0011)(1000)(1101)(0101). Now, this is equivalently 38D5 in hex.
783870
7x^2+16x over (x-2)(x+2)(x+3)
180224
Not necessarily.  If the answer to a division problem is 7.5, then the answer is 7 and a half (half of what you're dividnig by).\n\nExample- 15 divided by 2 is 7.5, but the remainder in this problem is 1, not 5 (1 is half of 2).
677314
To do this, you do (2/3) / (1/5). You must flip the denomiator fraction over and change the operation to multiplication. So... that would be (2/3) * (5/1) which would be 10/3.
562262
This is easy synthetic division, but you did not include a divisor. Anyway, you use each numbers co-efficients as the top numbers, leave a row, and then the bottom row, for the answer. The divisor must be x (variable) + or - (any number, say 1) \n(x+1) divide  (x^4 + x^3 + x^2 + x + 1).\ntake inverse of (x+1) integer: -1.\ndrop the first co-efficient and then divide co-efficients by this, 'bouncing' remainder into middle row. \n\nthe bottom row would be about 1/1/1/1 which you'd supplement on to the taken off variable, x.\nAbout This:\nx^3 +x^2 +x +1
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/
164299
Ask him/her for the remainders.\n\nHe/She soon discovers that in dividing by 5 we always have a remainder of 0,1,2,3 or 4. Not 5.\n\nAnd so on for other divisions.\n\nearl, This is my own experience from my 4th grade!\nNow that I am an amateur numbere theorist I know that remainders are more important than quotients.\nSo as a math instructor, this is my advice for \ndivision problems for a forth grader ;o)
775593
Load the Analysis Toolpak add-in and then use this formula, it's shorter:\n\n=A1/GCD(A1, B1)&":"&B1/GCD(A1, B1)
773850
Tip:  If the sum of the coefficients of the polynom are zero, then x =1 is a zero of the polynom.\n\nx = 1\nIf you need more roots do the synthetic divisiona and then use the quadratic formula.
890098
I enjoy algebra, and this is easily my favorite type of problem.  So I hope you, too, one day enjoy them for their inner beauty.\n\nFor now, let's tackle this "factoring problem."  We may wish to use the method of "synthetic division" which is arguably easier than "long division of polynomials."\n\nThink back to grade school where you learned that 6 = 2 * 3.  This simple statement enabled you to 'deconstruct' the number 6 as either "two times three" or "three times two".  Even if you forgot one of the 'factors' you could conjure it up again quickly by asking yourself, "What is six divided by two?" for example.\n\nIn problem (a), you are essentially asking, "What are the coefficients of P(x) if the only factor we know is (x-1)?"\n\nLet's set up synthetic division (see link below), by placing all the coefficients of P(x), and using 1 (the zero implied by (x-1)) as the divisor:\n\n1 ||  a  b  c  d\n___________\n.....a\n\n1 ||  a  b  c  d\n_____a_____\n.....a (a+b)\n\n1 ||  a  b        c       d\n_____a___(a+b)____\n.....a (a+b) (a+b+c)\n\n1 ||  a  b        c                d\n_____a___(a+b)___(a+b+c)_\n.....a (a+b) (a+b+c) (a+b+c+d)\n\nNote that this last term must equal 0 since it is assumed that (x-1) is a factor ... thus, the "remainder" term must equal zero!\n\nThere's the result you were after:  a+b+c+d = 0\n\nHope this helps you tackle the others!