Find all zeros of the polynomial  P(x) = x^4 - 3x^3 + 5x^2 - 27x - 36.?
408031
First, factor the polynomial using the Rational Roots Theorem. All possible factors of -36 are (+or-) 1, 2, 3, 4, 6, 9, 12, 18, 36.  Use synthetic division for each, until you find one root.  In my solution I got x - 4 as a root and the other x^3 + x^2 + 9x + 9.  Continue with the factoring.  Here I got x + 1 as a factor and the other x² + 9. Therefore, the factorization is:\nP(x) = (x - 4)(x + 1)(x² + 9)\n\nThe zeroes are all the roots when P(x) = 0\nTherefore,\n(x - 4)(x + 1)(x² + 9) = 0\nx = 4\nx = -1\nx² = -9 or \nx = 3i or -3i\n\n^_^
920343
(4/3)[x^3/(2-x)]^(1/3)\n[(2-x)(3)x^2-x^3(-1)]/\n(2-x)^2=\n\n 4x[1/(2-x)]^(1/3)\n[3x^2(2-x)+x^3]/\n3(2-x)^2
526420
u need to factor:\n\n(x^3-9x)=0\nx(x^2-9)=0\nx(x-3)(x+3)=0\nx=0//x-3=0//x+3=0\n      +3 +3  -3 -3\n     x=3    x=-3  \n\nso ur final answer is x= 3, -3\n\nhope this made sense!
950494
The rational zero theorem says that all possible rational zeros of a polynomial with integer coefficients, can be obtained by dividing the factors of the constant (in this case, it is -3) by the factors of the integer that is multiplied by the highest power of x (in this case, it it 2).\nAll factors of -3 are -1, +1, -3, +3.\nAll factors of 2 are -1, +1, -2, +2.\n\nThus, all possibilities are given by :\n-1/-1, -1/+1, -1/-2, -1/+2,\n+1/-1, +1/+1, +1/-2, +1/+2,\n-3/-1, -3/+1, -3/-2, -3/+2,\n+3/-1, +3/+1,+3/-2, +3/+2\n\nwhich leads to ±1, ±3, ±1/2, ±3/2 as the 8 possibilities.
1161859
3x^2 - x - 24 = 0\n(3x + 8)(x - 3) = 0\n\n3x + 8 = 0 and x - 3 = 0\nx  = -8/3 and x = 3
902478
wow its been a long time.. i use a calculator to do that now.. but here goes one thing i remember...\n\nlets say we have a polynomial like this\n\nx^2 -3x-10  and we need to factor it..\n\nwe take a look at what factors of -10 we could use that would add up to -3  in this case there are two way to get -10 (-2*5) or (2*-5)  if we add the two sets together one gives us -3 and the other gives us 3 so we can throw out the -2*5 terms.\n\nSo now we lay out the factors.... we need two x's to get x^2 so we need two x's in the first terms (remember FOIL?)\n\nlay the terms out so they give us the correct values\n\n(x+2)(x-5)  now we can use FOIL (firs, outer, inner, last) to check ourselves...\nX*X = x^2\nX*-5 = -5X\n2*x =2x\n2*-5=-10\n\nso we get [x^2] +[-5x] +[2x] +[-10]\nor x^2 -5x+2x-10\ncombine like terms now\n-5x+2x =-3x\n\nwhich gives us\nx^2-3x-10\n\nhope that helps
653904
1+3i is a zero means that the conjugate 1-3i is also a zero.\nThen x-(1+3i) and x-(1-3i) are factors of the polynomial.\nMultiply the two together.\n(x-(1+3i))(x-(1-3i)) = x^2 - x(1-3i) - x(1+3i) + (1+9)\n= x^2 - 2x + 10\nNow use long division.\n[x^4-3x^3+6x^2+2x-60]/[x^2 - 2x + 10]\n= x^2 - x - 6\nThis factors into (x-3)(x+2), so the other zeros are 3 and -2\nAnswer: 1+3i, 1-3i, 3, -2
1164377
There are zeros at 2, -3/2 and -6, so\n\nf(x) = (x-2)*(x+3/2)*(x+6)\n\nf(x) = x^3 + 11/2*x^2 - 6*x -18
831674
1. replace x with -3, calculate and see if you get 0\n2. divide by (x+3) and solve the remaining quadratic equation
983479
x^2-3x-2x+6=0\nx(x-3)-2(x-3)=0\n(x-2)(x-3)=0\nx=2 or 3
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"
925129
Let x = the number\nThen x^2-3x=54\nx^2-3x-54=0\n(x-9)(x+6)=0\nx-9=0\nx=9\nx+6=0\nx=-6\nx= 9 and x = -6 are the two possible answers
498152
To find the roots of a polynomial p=x^2 -2*x -5 in Matlab, user at the Matlab prompt >>, enters:\n\np = [1  -2 -5] ; please note that numbers in bracket represent the coefficient of the polynomial.\n\n \n\nTo that Matlab responds by:\n\n \n\np =     1     -2    -5\n\nThen user enters: r= roots(p).\n\nAt this point the roots are printed.
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.
875938
5\n\ntake the highest power in the polynomial, that exponent is your leading exponent, and is used as your "n"\n\nso 5 zeros.
117698
so that x=4+ or - 3i\nthen transpose 4 to left\nx-4=+/- 3i square both sides\nthat's x^2-8x+16=-9\ntranspose -9 to left side\nx^2-8x+25=0
86160
its simple.....answers \n1. x=-4, y=-3/4\n2. x=5,-1\n3. p=2,-1/2\n4. 4x^2-x+6\n\nif u want to know HOW to get these answers mail me sudeepjaiswal_2001@yahoo.com....actually im busy now...LOL\n\nseriously..mail me