This is a 2 part question about irreducible monic polynomials.?
402130
I get that the irreducible monics, with the primitive ones so marked, are:\n\nx^3 + 0 x^2 + 2 x + 1 (primitive)\nx^3 + 0 x^2 + 2 x + 2\nx^3 + 1 x^2 + 0 x + 2\nx^3 + 1 x^2 + 1 x + 2\nx^3 + 1 x^2 + 2 x + 1 (primitive)\nx^3 + 2 x^2 + 0 x + 1 (primitive)\nx^3 + 2 x^2 + 1 x + 1 (primitive)\nx^3 + 2 x^2 + 2 x + 2\n\nYour mileage may vary. :-)\n\nDegree three reducibles factor in one of the two forms:\n\n(x - a)(x - b)(x - c)\n(x - a)(x^2 + bx + c) with x^2 + bx + c irreducible\n\nSo if p is a degree three reducible then at least one of p(0), p(1), and p(2) must be 0 modulo 3.  Applying that test leaves the eight monic irreducibles shown above.\n\nThe primitive ones are found by verifying that x + <p(x)> has multiplicative order 26 in (Z/3Z)[x].
64159
It's debatable.  Here's an interesting column that addresses the question:\nhttp://mathforum.org/library/drmath/view/57188.html
221826
1) \nxy - x + 4y - 4\nx(y-1) + 4(y-1)\n(x+4)(y-1)\n\n2)note: x^3 means x cubed; x^2 means x squared\nx^3 -6x^2 + 2x - 12\nx^2(x-6) + 2(x-6)\n(x^2 + 2)(x-6)\n\n3)\n6tx + r - 3t - 2xr\n3t(2x - 1) - r(2x - 1)\n(3t - r)(2x - 1)\n\n4)Are you sure it wasn't -x^2 - 2x - xy - 2y?\n-x^2 - 2x - xy - 2y\n-x(x + 2) - y(x + 2)\n(-x - y)(x + 2)\n\nTo "factor by grouping," you must split the original polynomial that you are given into two different binomials by "grouping" them together.  This is done with perenthesis in the problems above.  Then, you factor out the "greatest common factor" from each of the binomials.  After factoring, the two quantities still within the perenthesis should be the same.  You could call these quantities the "common binomial."  Lastly, you factor out the common binomial to get the final answer.
333704
1. False, a polynomial has no = sign.\n2. False, the log of zero (or any negative number) does not exist.
78119
I guess you must factorize them.\nAll the polynomials you wrote are to be factorized by partial factorization.\nI think you know the general rule.\nA(x)P(x)+A(x)Q(x) + B(x)P(x) + B(x)Q(x) = \n---> factorize between the 1°-2° and 3°-4° <---\nA(x)[P(x) + Q(x)] + B(x)[P(x) + Q(x)]\nYou get the same binomial term [P(x) + Q(x)] which is a common factor. Factorize:\n[P(x) + Q(x)]*[A(x)+B(x)]\n\nFor example, in the first:\nA()=a\nB()=b\nP(x)=2x\nQ(y)=3y\nYou get \na(2x+3y)+b(2x+3y) = (2x+3y)(a+b)\n\nIn all the others you must rearrange the terms to use the above steps:\nin xy+32+8x+4y the repeating patterns are 8x + xy and 32 + 4y ....\n\nThink about it and good work
903397
You might try asking this question under "Mathematics."
565133
We start by factoring the polynomial.  A little trial and error (testing the factors of 2, which is the constant term) reveals that 2 is a root.  So, we know that (x-2) is a factor.\n\nDividing the polynomial by (x-2) using synthetic division, we find the other factor is 6x^2+x-1.  This can be factored as (3x-1) (2x+1).\n\nSo, the equation, in factored form, is\n(x-2)(3x-1)(2x+1)=0\n\nSo, the roots are x=2, x=1/3, and x=-1/2.
484948
A 3rd degree polynomial can have a maximum of 3 real roots, but it must have at least 1 real root I believe.
359378
This is the ring of polynomials in two variables with complex coefficients. First figure out what the units are (the scalars). An irreducible will be a polynomial p such that if p=fg, then either f or g is a unit. Clearly x and y are irreducible since if x=f(x,y)g(x,y), then either f or g will be a scalar polynomial.
621340
The first one is (x-2)(x+9)(x-7).\nHowever, for the second part, everyone else has forgotten that imaginary zeroes come in CONJUGATE PAIRS.\nSo there are actually 5 zeroes to part 2, 4i, -5, -2i, 2i, -4i.\n\nSo the solution would be (x-4i)(x+4i)(x+5)(x-2i)(x+2i)\n(x^2+16)(x+5)(x^2+4)\nx^5 + 5x^4 + 20x^3 +100x^2 + 64x + 320\nViola...
465550
Do you mean "zero of a polynomial?" If so, then read on...\n\nA zero of a polynomial is the value or values of the variable (usually x) that cause the polynomial to equal zero.\n\nExample 1: The zero of the monomial 3*x is 0 because 3*0 is zero.\n\nExample 2: The binomial x + 2 has it's zero at x = -2 because if you substitute -2 in place of x, then you get \n---> -2 + 2 = 0.\nTherefore, -2 is the zero of the given binomial.\n\nExample 3: Find the zeros of the trinomial x^2 - x - 6. That is, solve x^2 - x - 6 = 0. If you factor the trinomial, then you get \n---> (x + 2)(x - 3) = 0. \nIn this problem there are two possibilities. Either \n---> x + 2 = 0  or  x - 3 = 0. \nSolving each gives x = -2 or x = 3. Therefore, the zeros of the given trinomial are -2 and 3. \n\nIf you substitute -2 into the given trinomial, then you have the following:\nx^2 - x - 6 \n= (-2)^2 - (-2) - 6 \n= 4 + 2 - 6\n= 6 - 6\n= 0 \n\nIf you substitute 3 into the given trinomial, then you have the following:\nx^2 - x - 6 \n= (3)^2 - (3) - 6 \n= 9 - 3 - 6\n= 6 - 6\n= 0 \n\nThere are no other values of x that will cause our given trinomial to equal zero.\n\nDefinitions\nA monomial is a polynomial with one term.\nA binomial is a polynomial with two terms.\nA trinomial is a polynomial with three terms.\nA polynomial with more than three terms is just called a polynomial.\n\nNote: If you graph a polynomial, then the graph will touch or cross the x-axis at the zeros of the polynomial. Zeros of a polynomial are also called solutions, roots, and, in some cases, x-intercepts.
936881
There is a local maximum at x = c on f(x) if\nf'(c) = 0 AND f"(c) < 0\n\nOccasionally also when f"(c) = zero .. this occurs when there is a multiple root at that value of x.\nThen further testing is needed. In this case:\n1. if If f(c-) < f(c) AND f(c+)< f(c) then it is a local maximum;OR\n2. if f'(c-) >0 and f'(c+) <0 then it is a local maximum; OR\n3. if f"(c-) <0 and f"(c+) <0 then it is a local maximum
244218
http://mathworld.wolfram.com/GaloisGroup.html\n\nI'm not sure if I can give you the exact answer since I'm trying to understand it myself.  \nWhat you want is the Galois group of Q(r1, r2, r3, r4, r5) over Q where the ri are the roots to the polynomial x^5 + x + 1.  This group is going to be a subgroup of the group of permutations on 5, but I don't know if I can say anything else about it since x^5 + x + 1 is an irreducible polynomial in Q.  \nHope this helps.
175889
Proving irreducibility over Q is rather straightforward based on what the form of a nontrivial factorization would have to look like.\n\nReduction modulo every prime follows from X^4 - 10x^2 + 1 being equal to all three of the following:\n\nx^4 - 2x^2 + 1 - 8x^2 = (x^2 - 1)^2 - 2(2x)^2\nx^4 + 2x^2 + 1 - 12x^2 = (x^2 + 1)^2 - 3(2x)^2\nx^4 - 10x^2 + 25 - 24 = (x^2 - 5)^2 - 6(2^2)\n\nIf 2 is a quadratic residue modulo p, then the first is a difference of two squares.\n\nIf 3 is a quadratic residue modulo p, then the second is a difference of two squares.\n\nIf neither 2 nor 3 is a quadratic residue modulo p, then their product 6 is a quadratic residue modulo p, and so the third is a difference of two squares.
333000
We use the fact, by definition,\n[m] = 0 in Z(n) where,\nfor all m where m is divisible by n .\nand [m] = "remainder of m when divided by n".\n\n1) [21] + [19] in Z(12)\n\nSo, in Z(12),\n \n[21] + [19] =  [21 + 19]\n \n[21] + [19] =  [40]\n\n[21] + [19] =  [12*3 + 4]\n\n[21] + [19] =  [4] \n\n[21] + [19] =  4\n\n2) [34][27] in Z(45)\n\n[34][27] = [34*27]\n \n[34][27] = [3427]\n\n[34][27] = [3420 + 7]\n\n[34][27] = [45*76 + 7]\n\n[34][27] = [7]\n\n[34][27] = 7\n\n3) [33]([83]-[67]) in Z(100)\n\n[33]([83]-[67]) = [33]([83 - 67])\n\n[33]([83]-[67])  = [33] ([16])\n\n[33]([83]-[67])  = [33*16]\n\n[33]([83]-[67])  = [528]\n\n[33]([83]-[67])  = [5*100 + 28]\n\n[33]([83]-[67])  = [28]\n\n[33]([83]-[67])  = 28 \n     \n4) The additive inverse of [18] in Z(26)\n\nLet the additive inverse be y.\nBy definition [y] + [18] = 26 in Z(26)\n\nSo, \n[y] = 26 - [18]\n  \n[y] = [26] - [18]\n\n[y] = [26 - 18]\n\n[y] = [8]\n\nHence, y = 8.\n\nUse this to double - check your "homework" answers.\n\nRegards.
465556
Given that f(x) is a polynomial of degree EIGHT such that\n\nf(m) = 1/m for m = 1,2,3,4,5,...,9\n\nFind f(10)  \n\n\n*Answer: 2/10\n\n*Solution:\n\nConsider another polynomial, g, defined as g(x) = x f(x) - 1\n\ng(1)=0, g(2)=0, …, g(9)=0\n\nSo g(x) = A(x-1)(x-2)…(x-9)\n\ng(0) = 0 f(0) - 1 = -1\n\nalso, g(0) = A(-1)(-2)…(-9), so A = 1/9!\n\ng(10) = A(9)(8)…(1) = (1/9!)(9!) = 1\n\ng(10) = 1 = 10 f(10) - 1, so f(10) = 2/10\n\n\n**Or try doing some research on "Rabin's method"
739088
Consider a = 1, b = 0, c = 1.  Clearly, b^2-4ac = -4 < 0.  This corresponds to the equation x^2 + 1 = 0.  But x^2 is never negative, and 1 more than that is always positive, so this has no solutions.\n\nIf you want only non-imaginaries, there are no solutions when b^2 < 4ac and only one solution when b^2 = 4ac.\n\nYou clarify: "that does not suffice because every quadratic is meant to give probable solutions because all polynomials and quotients of polynomials are rational, and rational equations don't give imaginary solutions."\n\nRational equations do indeed give complex solutions, many of them nonreal.  The example I gave above (x^2 + 1 = 0) is an example.  The algebraic closure of the reals is the complex numbers; the algebraic closure of the rationals is the algebraic numbers.  If you want all quadratics to give two solutions (counting multiplicity), you'll need to allow imaginaries.