Finding Zeros. Find all the zeros  of the polynomial function.?
602459
How do you expect to learn something if you get others to do your homework.  If you do not know how to do something, you should go to your teacher and ask him to explain it further to you.
774094
- Find the x-intercepts, by setting y=0 and solving the polynomial equation. You may need to factorise\n\n- Find the y-intercepts, by setting x=0 and solving.\n\n- Find stationary points by differentiating the polynomial equation wrt x and setting the differential to 0. These give you the stationary points.\n\n- Differentiate again to find out whether these turning points are maxima, minima or points of inflexion:\nIf d²y/dx²  is positive, then it is a minimum point.  \nIf d²y/dx² is negative, then it is a maximum point  \nIf d²y/dx² is zero, then it could be a max, a min or a point of inflexion.\n\n- Find asymptotes. You may get a horizontal asymptote when x approaches +ve or -ve infinity and a vertical one as y approaches +ve,-ve infinity.\n\nThis is the basic info you need to sketch a graph depending on how complex the polynomial is.
527455
If those are the zeros, you just do it like this for the first one:\n\ny=(x-5)(x-3i)(x+3i)\n\n\nIf your teacher wants it in general form, you have to multiply out all the terms. \n\ny=(x-5)(x^2+9)\n\ny=x^3+4x^2-45\n\n\nThe second set doesn't make sense because you have an i without a partner. In polynomial equations, the complex factors always come as conjugate pairs, like the -3i and +3i.\n\nAnyway, I hope this helps. Basically, you set up a bunch of factors that are x-each root.\n\nI just changed it because I realized that the 1. meant problem number 1.
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^_^
923617
First, factor out 3x from the equation, so 0= 3x(x^2-2x-3).\n\nThen, factor the polynomial in the parenthesis, so 0= 3x(x-3)(x+1)\n\nAfter that, use the zero product property to set each part of the the equation to zero, so 0=3x, 0=x-3, and 0=x+1.\n\nSolve these three to get x=0, x=3, and x=-1, respectively.
220199
The problem factors to\n\n(x - 2)(x^2 + 1)(x^2 + 1) or (x - 2)(x^2 + 1)^2\n\nx = 2, i, -i\n\nthe only real zero is 2.
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 ;-)
846578
1.  Since f(x) = x² + bx + c, and you know f(4) = 9 (since (4,9) is a point on the parabola, specifically the vertex), you can solve for f(4) = \n\n16 + 4b + c = 9, so 4b + c = -7\n\nSince (4,9) is the vertex, you know the tangent at this point is zero, i.e., f '(4) = 0.  (I hope this is a calculus HW problem so that you know what I am talking about!)\n\nSo get f '(x) = 2x + b\n\nSince f '(4) = 0, we have 2(4) + b = 0, so b = -8\n\nSince b = -8, and 4b + c = -7, we can now solve for c:\n\n4(-8) + c = -7, so c = 25\n\n----------\n\n2.  Not sure I can help here.  I took the first and second derivatives of h(x) to determine that the function has a local min at x=7/3 and a local max at x=-1, but h(7/3) ≠ -5 and h(-1) ≠ 2, so not sure where to go from here, sorry!\n\n----------\n\n3.  A parabola has a 2nd-degree polynomial equation, or y = ax² + bx + c.  We know two values, namely (4,4) and (2,8) so we can set up two equations like thus:\n\n4 = a(4²) + b(4) + c\n8 = a(2²) + b(2) + c\n\nwhich we can rewrite as:\n\n4 = 16a + 4b + c\n8 = 4a + 2b +c\n\nRight now we have 3 unknowns and only 2 equations, but we can get a third equation by taking the first derivative, or f '(x) = 2ax + b.  Since the vertex is at (4,4) we know that f '(4) = 0, and we can get a third equation:\n\n0 = 2a(4) + b = 8a + b\n\nSo now we have 3 equations in 3 unknowns and we can solve for a, b, and c:\n\n4 = 16a + 4b + c\n8 = 4a + 2b +c\n0 = 8a + b\n\nSubtract equation 1 from equation 2, then divide by 2 to get:  \n2 = -6a - b\n\nNow add equation 3 to this to get 2 = 2a, or a=1.\n\nNow let's back-substitute to get b and c:\n\n2 = -6(1) - b, so b = -8\n\n8 = 4a + 2b +c\n8 = 4(1) + 2(-8) + c, so c = 20\n\nSo the equation of our parabola is:\nf(x) = x² -8x + 20\n\n(Double check to make sure f(4)=4 and f(2)=8).
762440
Just split it into parts - when is x-x^2<0, and when is it >0?\nThat bit has roots 0 and 1, so it is positive between 0 and 1, negative everywhere else.\n\nThus, ∫(x=-1 to x=0) (x^2-x) dx + ∫(x=0 to x=1) (x-x^2) dx + ∫(x=1 to x=3) (x^2 - x) dx\n\n= [(1/3 x^3 - 1/2 x^2) from -1 to 0] + [(1/2 x^2 - 1/3 x^3) from 0 to 1] + [(1/3 x^3 - 1/2 x^2) from 1 to 3]\n= (0- -5/6) + (1/6 - 0) + (9/2 - -1/6)\n= 17/3.
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.
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...
895342
x^2-9x-x+9=0\n(x-9)(x-1)=0\nx=9 or x=1
766816
This problem is best solved in a manner identical to how all other polynomials are solved. So x^2=x^3? This is the same as x^3-x^2=0, so factoring this yields x times x times (x-1)=0 so the only possible solutions are x=0 and x=1.
774014
2^3+2*2^2-3(2)-10=8+8- 6 -10 =0\nTherefore x-2 is a factor of the given polynomial
759615
The easiest function of x are:\n\nf(x) = constant\nf(x) = a*x, where a is a constant : 1st order poly\nf(x) = b*x^2 where b is a constant 2nd order\nf(x) = c*x^3 where c is a constant: 3rd order\n...\nf(x) = z*x^n, where z is a constant, n is a positive integer, nth order polynomial\n\na general polynomial is a sum of these.  They are convenient to work with for lots of reasons.  That's why they are so important.
784758
x^2+x+1=0\nx^2+2(1/2)x+1=0\nX^2+2(1/2)x+(1/2)^2+1-(1/2)^2=0\n(x+1/2)^2+3/4=0\n(x+1/2)^2=-3/4\nx=-1/2+/-√(3/4) i=-1/2+/-√3 * i/2
60538
if [ -7x + x^2 =0 ]\nthe you will get y = 7/x