Using the remainder theorem... question?
139386
(X-c) here represents the number you are going to devide with. c can take any value here.\nLet you have to devide with (x-5) , then c will be 5.\nAnd if you have to devide with (x+10) then c will be -10. As (x-(-10))=(x+10).\nThe remaider therum says, If you devide a polynomial, P(x) by (x-c) then the remainder will be P(c).\nIf you want a proof of this theorum,\nlet, the divisor of divison of P(x) by (x-c) is A and remainder is B.\nSo P(x)=A*(x-c)+B.\nSo P(c)=A*(c-c)+b\nor, B=P(c) which is the remainder, I hope this will clear your concept.
1030587
I like to use C(n,k) for the combiniation of n things taken k at a time.\nUse the binomial theorem to expand (1+k)^n\n\n(1+k)^n = C(n,0) (1^n)k^0 + C(n,1)(1^(n-1))k^1 + C(n,2)(1^(n-2))k^2 + ... + C(n,n-1)(1^1)k^(n-1) + C(n,n)(1^0)k^n\n\n1 ^ of anything is 1 , and if we set k=1, then we get\n\n(1+1)^n = 2^n = C(n,0) + C(n,1) + C(n,2) + ... + C(n,n-1) + C(n,n)\n\nif this sum =2048, then 2^n=2048 or n=11
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
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)
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).
1177677
shouldn't it be x^2??? if it IS x-squared, then sure, you could use the factorization method. \n\nit goes like this:  (6x + 1) (x - 1) = 0\n\ntherefore, x = -1/6 , 1
5300
(n+k-1) over (k) = (n+k-1) / (k)\n\n=> (n+k-1) / (k) = n/k +1 -1/k
473264
all factorials greater than 6! divide evenly since they contain 7\nso add 1! to 6! and divide by 7, find remainder\n=1!+2!+3!+4!+5!+6!\n=1+2+6+24+120+720\n=873/7\n=126 R 5\n\nanswer is 5
561262
u jammed the equation together.\ndo u mean\n6x^3 + ( 6/x - 2 * x^2 - x/3x2 - 3x + 3)\nits really hard to see ur question. can u write it out with spaces...\nhurry up i am waiting... lol
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!
38332
(x-2)/(x^2-2x-15)+(x+2)/(5-x)\n(x-2)/((x+3)(x-5))+(x+2)/(5-x)\n(x-2)/((x+3)(x-5))+(-2-x)(x+3)/(x-5)(x+3)\n\n[(x-2) + (x+3)(-2-x)] / [(x+3)(x-5)]\n\nANSWER :\n(x^2+4x+8)/((x+3)(5-x))
773695
Look in your text book under "factoring polynomials"... or you could try this web site that explains this fairly thoroughly...\n\nhttp://faculty.ed.umuc.edu/~swalsh/Math%20Articles/Factoring.html
1065775
If a(N)=Sum(r^n, 0=>N) and -1<r<1...\n\n Is something that looks like 1/(1-r^N)
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.
51652
Remainder and modulo are really the same concept mathematically.  Remainder is the integer amount left over after dividing the denominator into the numerator as many whole number times as you can.\n\n2 goes into 5, 2 times.  (4 is divisible by 2 and is the largest whole number that is less than or equal to 5).\n\n5 / 2 remainder is 1.\n5 modulo 2 is 1.\n5 % 2 is 1.\n\nOne way of expressing this in some programming languages is:\n\na = 5, b = 2\nremainder = b * ((a / b) - int(a / b))\nremainder = 2 * ((5 / 2) - int(5 / 2))\nremainder = 2 * ((2.5) - int(2.5))\nremainder = 2 * ((2.5) - (2))\nremainder = 2 * (0.5)\nremainder = 1
1053054
Use the binomial.  I've already shown you how in a previous question.  And you already know that you need to use it.
228982
You  need to show that 2^1386 has remainder 1 when divided by 1387. So work modulo 1387 and obtain\n\n2^18=262144=1 (mod 1387).\n\nNow, 1386=18*77, so 2^1386=(2^18)^77=1^77=1 \n(mod 1387).