Q is the rational numbers.
123814
The roots are of the form root(8,3)*omega^j where omega is a primitive 8th root of unity. Now, we can independently send root(8,3) to any root and omega to any primitive root of unity. There are four such primitive roots, so the group will have 8*4=32 elements. The subgroup that leaves root(8,3) fixed will be a Klein 4-group, while the subgroup leaving omega fixed will be cyclic of order 8. You go from there.
349652
(x^4) + (4x^3) + (2x^2) - 8x - 8=0\n\nGeneral:  ax^4+bx^3+cx^2+dx+e=0\nif p/q is a rational root then p is divisor in a and q is divisor in e =>\n\nin this case:\np belong to {-1,+1} & q belongs to {-8,-4,-2,-1,1,2,4,8} =>\n\np/q belong to {-8,-4,-2,-1,1,2,4,8}\n\nWe try  -2 as a root and see\n2^4 + 4*2^3 + 2 * 2^2 - 8 * 2 - 8=0\n\nWe can now do polynomial division by (x+2):\n\n(x^4) + (4x^3) + (2x^2) - 8x - 8 : (x+2) = x^3 +2x^2 -2x - 4\nx^4 . . . + 2x^3\n- - - - - - - - - -\n0 . . . . . . 2x^3\n. . . . . . . .2x^3 . .+4x^2\n- - - - - - - - - - - - - - - \n. . . . . . . . .0. . . -2x^2 . . -8x\n. . . . . . . . .  . . . -2x^2 . . -4x\n- - - - - - - - - - - - - - - - - - - -\n. . . . . . . . . . . . . . . . . . .-4x. -8\n. . . .  . . . . . . . . . . . . . . -4x  -8\n- - - - - - - - - - - - - - - - - - - - - - -\n. . . . . . . . . . . . . . . . . . . . . .  0\n\nexactly the same procedure as before gives x= -2 is root so now we go again\n\nx^3 +2x^2 -2x - 4 : (x+2) = x^2-2\nx3^. .2x^2\n- - - - - - -  - -\n. . . . . . .0 . .-2x -4\n. . . . . . .0 .  -2x -4\n- - - - - - - - - - - - -\n. . . . . . . . . . . . .0\n\nthe last term x^2-2=0 => x=±√ 2\n\nnow we have :\n\n(x^4) + (4x^3) + (2x^2) - 8x - 8 =\n(x+2)^2 * (x-√ 2) * (x+√ 2) = 0\n=======================\nso x = { x in R I x = -2 or x= -√ 2 or x= +√ 2}\nand x= -2 is root 2 time\n=================================
429588
There is no shorter way than, first, to tabulate n(n+1) mod 5 for each possible value of n :-\n\nn = 0, n(n+1) = 0 * 1 = 0\nn = 1, n(n+1) = 1 * 2 = 2\nn = 2, n(n+1) = 2 * 3 = 6\nn = 3, n(n+1) = 3 * 4 = 12 = 2 mod 5\nn = 4, n(n+1) = 4 * 0 = 0\n\nNote that 0 and 2 can be halved, and we have not reduced 6 to 1 because we are going to halve it too.\n\nSo half of n(n+1) can only be 0, 1 or 3.  QED.
586154
no:\nsqrt(2) + (-sqrt(2)) = 0 rational\nno\nsqrt(2)*(1/sqrt(2)) = 1 rational\n\nsqrt(2) and 1/sqrt(2) are irrational\n\n In general , it's more often that sum ( product) of too irrational is irrational
428548
You have done some mistake while copying the problem.Please check it up.It may be 36l^2-36lq+9q^2 or something else.However I am doing ipresuming it to be 36l^2-36lq+9q^2\n\ngiven exp.=9(4l^2-4lq+q^2)\n=9{2l)^2-2.2l.q+(q)^2}\n=9(2l-q)^2\nThe answer may also be written as {3(2l-q)}^2 or (6l-3q)^2
156188
One way is to construct the rational numbers so that each is written as a fraction that does not need to be simplified.  \n\nBegin with all fractions whose numerator and denomenator sum to 2.  There is only one such fraction (1/1).  Map this fraction to the integer 1\n\n1/1 --> 1\n\nThen consider all fractions whose numberator and denomenator sum to 3.  There are two such fractions:  2/1 and 1/2.  Map these two fractions to the next available integers:\n\n2/1 --> 2\n1/2 --> 3\n\nThen map the three fractions whose numerators and denominators sum to 4 to the next available integers:\n\n3/1 --> 4\n2/2 --> 5\n1/3 --> 6\n\nAnd so forth.\n\nThis algorithm will map all of the rational numbers onto the set of natural numbers.  However, it is a many-to-1, not a 1-to-1 mapping (it maps 1/1 to 1 and also maps 2/2 to 5, but 1/1 and 2/2 have the same value as rational numbers).  You could get around this by skipping over rational numbers that are not simplified fractions.
303350
same reason the "q" and "a" are.
229292
a. (x+9)(x-9)\nb.(t-2)(t+2)\nc. (b-12)(b+12)
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a. X x 3\n\nb. X / 3\n\nAnswers:\na. 19683\nb. .0018289895 or 4/2187
658266
The dist is\nsqrt.[(4-3)^2+(1-1)^2+(4-3)^2]=sqrt.2
284671
possible rational factors are p/q where, p is factors of the constant and q is factors of the leading coefficient.\np = 15 and q = 2\nfactors of p: 1, 3, 5, 15\nfactors of q: 1,2\nPossible p/q = (plus or minus) 1, 3, 5, 15, 1/2, 3/2, 5/2, and 15/2\nKeep picking one of these values and test it until you get a remainder of 0.\n1 works\n1 2 3 -11 -9 15\n ___ 2  5  -6 -15\n__2 5 -6 -15 0\nThus,\n2x^4 + 3x^3 - 11x^2 - 9x + 15 \n= (x-1)(2x^3 + 5x^2 - 6x -15)\n2x^3 + 5x^2 - 6x -15 can be factored by grouping.\n(2x^3 + 5x^2) + (- 6x -15)\n= x^2(2x+5) -3(2x+5)\n= (x^2-3)(2x+5)\nAnswer: (x-1)(x^2-3)(2x+5)
584571
R'(q) = 5(4q)/(2q^2 + 5)\nR'(4) = 5(16)/(32+5)\nR'(4) = 80/37
490767
a Q1 since it is between 0 and 67\nb Q2 since it is between 67 and 88\nc Q3 since it is between 88 and 105
538269
there are 3 cases for a number A raised to the power of n:\n\n1) n greater than or equals to 1.\nWe simply say that "A is raised to the power of n"\n\n2) 0 < n < 1 and n is a rational fraction expressed as 1/m, m is an integer\nWe say that "A is raised to the mth root"\n\n3) n is any rational fraction not included in case 2, expressed as p/q, p and q are integers\nWe say that "A is raised to the power of p and qth root" which therefore involves both power and root\n\nAnd you're right about radicals being the sign used. By right, for square roots (which is actually 2nd root), we should add a "2" on the top left hand corner of the radical sign, but by default a radical without any number is taken as square root. But for all other nth roots, the radical sign must be accompanied by the number n.
562319
square root of 13 is the answer
395786
Notice that both terms have a 'factor' of (p+q) \nThe first term is (p+q)*(p+q) and the second is (5)*(p+q)\nSo 'factor out' the (p+q) from each :\n= (p+q) * [ (p+q) + (5) ]\n= (p+q) * (p+q+5)\nis the factorization.
253607
Yes. the RHS of the first is equivalent to xor(P,Q). This becomes an inclusive or when the or(Q,...) is added. Not(P) as suggested above doesn't give the correct result when Q and P both = true.