Find the polynomial of least degree that has the given zeros.     x=-2, x=5i?
543826
(x+2)*(x-5i) = x^2+(2-5i)x-10i\n\nsomehow, i dont like these complex #'s\n\nyeah, the answer below mine is probably correct.\n\n(x+2)*(x-5i)*(x+5i) = ...
771554
(x+5)(x-5)\n(x)^2-(5)^2\n=x^2-25
898700
x = 4 ==> (x - 4) = 0\nx = -2 ==> (x + 2) = 0\n(x - 4)(x + 2) = 0\nx^2 - 2x - 8 = 0
778985
The rational zeros will be of the form p/q where p is a factor of 3 (the lowest order term) and q is a factor of 2 (the coefficient of the highest order x term, in this case 2x^3).  That means the possible answers are +/- 3/1, 3/2, 1/1 and 1/2.  Just plug those in to see which works.  (1, -1, 1.5)
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.
1027133
x^2-9x-10=0\n\nx^2 +(-1 +10)x -10=0\n\nx^2 -x +10x -10=0\nx(x-1) +10(x-1)=0\n\n(x+10)(x-1)=0\nthus x=10 , -1
572493
=x(x^3-2)\n--> x=0 or x=2^(1/3) or -2^(1/3)
851434
-x^2 if x=-2\n-(-2)^2  \n-(4)\n-4
738884
x^2-5x-14=0\n=>x^2-7x+2x-14=0\n=>x(x-7)+2(x-7)=0\n=>(x-7)(x+2)=0\nTherefore either x-7=0 or x+2=0\nIf x-7=0,then x=7,and if x+2=0,then x=_2\nThereforex=7 or -2 ans
948937
(-3,-3);(-2,0);(-1,1);\n(0,0);(1,-3)
1167006
If the zeros are: -4, 0, and 2, our factored equation must look like this:\nf(x) = x(x+4)(x-2)\n\nFoil it out:\n\nf(x) = x(x2 + 4x - 2x - 8)\nf(x) = x(x2 + 2x -8)\n\nf(x) = x3 + 2x2 -8x\n\nAnswer number 3.\n\nHope that helps.
852832
x = [-b +/- sqrt(b^2 - 4ac)] / (2a)\nf(x): a = 1, b = 0, c = -4\ng(x): a = 1, b = -1, c = -12
694939
Your parabola will be of the form f(x) = k*(x-int1)(x-int2) where int1 and int 2 are the intercepts given and then you adjust k so that f(x) passes through that minimum.  So f(x) = k*(x+2)*(x-3).  \n\nThe minimum is going to happen half way between the two intercepts, or at x=.5, and f(.5) = k*2.5*(-2.5) = -6.25k which is supposed to be -5 from the problem.  So then k must be .8 (since .8 * -6.25 = -5) and your polynomial is:\n\nf(x) = .8(x+2)(x-3) = .8x^2 - .8x - 4.8
782446
x:-    0** 2** 4\ny:-  -4** 0** 4\n\ni hope that this helps
207106
x^2-3=-2\nx^2=-2+3\nx^2= 1\nx=1\n\n1^2+-3+2=0
60538
if [ -7x + x^2 =0 ]\nthe you will get y = 7/x
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.