Find a polynomial of degree 4 with the following zeros: i, 3 + i?
917843
(x + i)(x - i)(x - 3 + i)(x - 3 - i) = 0\n\n(x^2 - i^2)[(x - 3)^2 - i^2] = 0\n\n(x^2 + 1)(x^2 - 6x + 9 + 1) = 0\n\nx^4 - 6x^3 + 10x^2 + x^2 - 6x + 10 = 0\n\nx^4 - 6x^3 + 11x^2 - 6x + 10 = 0   (Answer)
122155
(x + 1)(x + 3)(x - 2) = 0\n(x^2 + 4x + 3)(x - 2) = 0\n(x^2)(x - 2) + 4x(x - 2) + 3(x - 2) = 0\nx^3 - 2x^2 + 4x^2 - 8x + 3x - 6 = 0\nx^3 + 2x^2 - 5x - 6 = 0 is the answer ;-)
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"
559973
4i is the answer you seek,\nthe i is the square of -1 says any geek.\nThis is a abstract answer to solve your query,\nbut all negative roots include the 'i,' my deary.
877357
#1\n\nt=0 or -3\n\n#2\n\np^2 - (1/4)^2 which is p^2 - 1/16
866735
I believe you use the remainder theorem for this, which states that the remainder when a polynomial f(x) is divided by (x-k) is\nf(k)\n\nThus the remainder would be f(-1) = 11
863763
D ( 2 or 0)\n\nThis is because the coefficients of the given cubic polynomial are real numbers and so any real zero needs be accompanied by its complex conjugate and so as stated  earlier MUST come in pairs so either there are 2 or there cannot be any.
936498
let p(x) be given polynomial\n\nwe reallize that sqrt(3)/2 = cos pi/6\n\nso 4x^3-3x + cos pi/ 6= 0\n\nor - cos  pi/6 =-3x + 4x^3 = cos(3t) if x= cos t\n\nso cos 3t = - cos pi/6 = cos 5pi/6\n\n t = 5pi/18 or cos (2pi/3-5pi/18) = cos(7pi/18) \n\n or cos 17pi/18\n\nso a = 4\n\nb= cos 5pi/18\nc= cos 7pi/18\nd = cos 17pi/18\nas these are zeroes of polynomial \nPlease check the calculation steps
670366
Yes.  There are algorithms for factoring polynomials up to 4th degree.  These algorithms are guaranteed to work, however in the case of the 5th degree polynomials and above, there is no algorithm that is guaranteed to find the solution.  In part this is because the roots can be irrational numbers.  I was proved by Abel in 1824 that no such algorithm can even exist.\n\nThere are iterative methods to factor polynomials of 5th degree and above, but they provide approximations, they are not guaranteed to find the correct answer.  The iterative methods are quite complicated and are equivalent to finding eigenvalues of matrices.\n\nIf you are in high school or early college, all the problems they will assign you should be solvable with either the linear, quadratic, or cubic equations.
909578
Since the coefficient of x^4 is one, the possible rational zeros are just the factors of 8.\nList all the factors of 8, both the negative and positive.\nAnswer: -1, +1, -2, +2, -4, +4, -8, +8
661698
2b^4(4-3b)=0\n\nb=0\nb= (4/3)
593998
When you have a polynomial, the coefficients are integers.  For example:\n\nax^2 + bx + c = 0\n\nThis is the general formula for a second degree polynomial (x is to the 2, hence second degree).  In this case, a, b, and c are the integers.\n\nAlso, if you add other terms, such as x^3, they would also be multiplied by integers.
337374
i^2 = -1\nSo, i^4 = 1.\nSo, i^4n = 1.\n\nSo, i^335 = i^(4*83+3) = i^(4*83)*i^3 = 1*(-1)*i = -i
216863
((4 * 4) / 4) - 4 + 4 = 4
361536
Let's call f(x) = x^2 - 4\n\nThe graph of f(x) = x^2 - 4 looks like a smile ( U )  with the graph passing the y - axis at -4.\n\nThe roots are at x = 2 and x = -2.  This can be found by solving for x^2 - 4 = 0\nx^2 - 4 = 0\n(x + 2)(x - 2) = 0\nso x + 2 = 0, or x - 2 = 0\nso x = -2 or x = 2\n\nsince you're int'ing from x = -1 to x = 3, you have to break up the int into\n[x = -1 to x = 2, where f(x) is negative and take the absolute value] \nand\n[x = 2 to x = 3, where f(x) is positive]\n\nso int of f(x)\n= x^3/3 - 4x + c\n\nso int of f(x) from x=-1 to x=3\n= {+ve value of int of f(x) from x = -1 to 2}\n                +\n  int of f(x) from x = 2 to 3\n\nSeparately\nint of f(x), x = -1 to 2\n= | (2^3)/3  - 4(2) - [((-1)^3)/3  - 4(-1)] |\n= | (8/3) - 8 - [(-1/3) + 4] |\n= | (8/3) - 8 + (1/3) - 4 |\n= | (9/3) - 12 |\n= | 3 - 12 |\n= | -9 |\n= 9\n\nint f(x), x = 2 to 3\n= (3^3)/3  - 4(3) - [(2^3)/3  - 4(2)]\n= 27/3 - 12 - [(8/3) - 8]\n= 9 - 12 - (8/3) + 8\n= 5 - (8/3)\n= 7/3 (or 2 1/3)\n\nAdd them together and you get:\n9 + 2 1/3\n= 11 1/3 which is appromixately 11.333_
799253
Polynomial is a general term for any expression, including monomial, binomial, trinomial, etc.  Monomials have just one term with no + or -, such as x or 3x or 5x^6.  Binomials are two monomials added or subtracted, such as 3x + 5 or 5x^2 - 6x.  The degree of a polynomial is the highest exponent of any of its variables, assuming there is only one variable.  So for x^3 + 8x^2 + 3x - 4, the degree is 3.
230941
Write the coefficients of the variables\n-4__ 1 -3 -7 -31\n_____-4 28 -84\n____1 -7 21 -115\nSynthetic division is hard to type.\nThe remainder is the last number in the bottomw row, so\nP(-4) = -115\nTo check, just evaluate the polynomial P(x) at -4.  That is, replace all the x's with -4\nP(-4) = (-4)^3 - 3(-4)^2 - 7(-4) -31\n= -64 - 48 + 28 - 31\n= -115\nThus, this verifies that the correct answer is -115.