what is the fixed polynomial..i can't get it. The math problem is 3x=(6x squared)?
336950
3x = 6x^2 if rearranged becomes\n\n0 = 6x^2 - 3x\n6x^2 - 3x = 0\n\nsince the highest common factor for both terms is 3x, then\n6x^2 - 3x = 0\n3x(2x - 1) = 0\n\nremember that a x 0 = 0 x a = 0\n\nso either 3x = 0    or    (2x - 1) = 0\n\nif 3x = 0, then x = 0\nif 2x - 1 = 0, then 2x = 1 and x = 1/2
819891
2(x-6)=3x\n\n1.Distribute the 2 to the X and the 6. \n2x-12=3x\n\n2. Subtract the 2x from the left side and subtract it from 3x\n-12 = 1x\n\n3.flip the problem so it looks pretty!\nx = -12\n\nYour Welcome :D
1032854
given the zeroes of y:\nif x = -4, 0 or 2, then y = 0.  Therefore it passes through\n(-4, 0)\n(0,0)\n(2,0)\n\nIn other words -4, 0 and 2 are x-intercepts of the polynomial.\n\nThe polynomial (3rd degree) can be expressed as:\ny = x³ + ax² + bx + c\nIt can also be expressed as\ny = (x - x1)(x - x2)(x - x3) \nwhere n is a constant.\n\nWhen you substitute x1 to the expression\ny = (x1 - x1)(x1 - x2)(x1 - x3)\ny = (0)(x1 - x2)(x1 - x3)\ny = 0\n\nTherefore, x1 is a zero\n\nThe same goes with x2 and x3. Therefore you can put the zeroes into our expression\n\ny = (x - (-4))(x - 0)(x - 2)\ny = (x + 4)(x)(x - 2)\ny = x(x + 4)(x - 2)\nor\ny = x(x² + 2x - 8)\ny = x³ + 2x² - 8x\n\nThat is your 3rd degree polynomial! Hope this helps^_^
888367
You need to give some directions.  What is the problem asking you to do?\nNote: The person below is incorrect.  (4+3x)^5 does NOT equal\n4^5 + (3x)^5
941571
1) f(x) = A(x-1)(x-2)(x-4) for some A. Substitute in 0 to get -16 = A*-1*-2*-4, so A = 2. So 2(x-1)(x-2)(x-4).\n2) Whenever you have a polynomial with real coefficients and a+bi is a root, you also have a-bi is a root. So you know the 3 roots are 4, 3+i, 3-i. Thus f(x) = (x-4)(x-(3+i))(x-(3-i)). Multiplying those last two together, you get x^2 - x(-3-i-3+i) + (3+i)(3-i) = x^2 - 6x + (9-i^2) = x^2-6x+10. So f(x) = (x-4)(x^2-6x+10).\n3) I don't think you wrote the question down properly. Expanding the left hand side gives 3x^3 + (-a-27)x^2 + (9a+42)x - 14a. You need a=5 for the first part to work, but that doesn't give what you wrote.
723914
Are you certain the denominator isn't:\n\nx^6 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6x +1 ?\n\nBecause that denominator has a very simple factorization: (x-1)^6. Whereas the polynomial you actually gave for the denominator doesn't have any factors I'd be inclined to call simple, and I'll bet even money that there's no way of expressing them using any finite sequence of elementary operations and nth roots. I think you should double-check your input.\n\nEdit: well, it's a very good thing nobody took me up on that bet - it seems I am completely wrong. See the question at http://answers.yahoo.com/question/index;_ylt=Aqz3W3s3NPcz.j__CQYxK0bsy6IX?qid=20061127003230AA2oHuA\n\nI will give a summary here:\n\nAs mentioned, (x-1)^6 = x^6 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6x +1. So logically (assuming that really isn't a typo), your original denominator is (x-1)^6 - 10x^3. We need to find the roots of this equation, so set:\n\n(x-1)^6 - 10x^3 = 0\n(x-1)^6 - 10√x^6 = 0\nSince x=0 is not a solution, we may divide by √x^6. so:\n(√x-1/√x)^6 - 10 = 0\n\nLet t=√x-1/√x. Then this polynomial is:\n\nt^6-10=0\n\nThe solutions of this polynomial are precisely the complex sixth roots of 10, which are:\n\nt=√(∛(10))\nt=√(∛(10)) * (1/2 + i√3/2)\nt=√(∛(10)) * (-1/2 + i√3/2)\nt=-√(∛(10))\nt=√(∛(10)) * (-1/2 - i√3/2)\nt=√(∛(10)) * (-1/2 + i√3/2)\n\nWe now set √x-1/√x equal to each of these roots in turn. In so doing, we find the possible roots of x. Let u=√x. Then:\n\nu-1/u=t\nu²-tu-1=0\nu = (t±√(t²+4))/2\nx = u² = (t² ± t√(t²+4) + 2)/2\n\nx=(∛10 ± √(∛100+4∛10) + 2)/2\n\nx=(∛10 * (-1/2 + i√3/2) ± √(∛(10)) * (1/2 + i√3/2)√(∛10 * (-1/2 + i√3/2) + 4) + 2)/2 \n= (∛10 * (-1/2 + i√3/2) ± √(∛100 * (-1/2 - i√3/2) + 4∛10 * (-1/2 + i√3/2)) + 2)/2 \n= (∛10 * (-1/2 + i√3/2) ± √(-(∛100 + 4∛10)/2 + (4∛10 - ∛100)i√3/2) + 2)/2\n\nx=(∛10 * (-1/2 - i√3/2) ± √(∛(10)) * (-1/2 + i√3/2)√(∛10 * (-1/2 - i√3/2) + 4) + 2)/2 \n= (∛10 * (-1/2 - i√3/2) ± √(∛100 * (-1/2 + i√3/2) + 4∛10 * (-1/2 - i√3/2)) + 2)/2 \n= (∛10 * (-1/2 - i√3/2) ± √(-(∛100 + 4∛10)/2 + (∛100 - 4∛10)i√3/2) + 2)/2\n\nThus the factors of the denominator are:\n\n(x - (∛10 + √(∛100+4∛10) + 2)/2) * (x - (∛10 - √(∛100+4∛10) + 2)/2) * (x - (∛10 * (-1/2 + i√3/2) + √(-(∛100 + 4∛10)/2 + (4∛10 - ∛100)i√3/2) + 2)/2) * (x - (∛10 * (-1/2 + i√3/2) - √(-(∛100 + 4∛10)/2 + (4∛10 - ∛100)i√3/2) + 2)/2) * (x - (∛10 * (-1/2 - i√3/2) + √(-(∛100 + 4∛10)/2 + (∛100 - 4∛10)i√3/2) + 2)/2) * (x - (∛10 * (-1/2 - i√3/2) - √(-(∛100 + 4∛10)/2 + (∛100 - 4∛10)i√3/2) + 2)/2).\n\nAnd yes, that is one ugly polynomial.
455652
By the remainder theorem, substituting a value b into a polynomial gives the remainder if you divided the polynomial by (x - b). So if you can find a value that goes into the polynomial that gives a value of 0, then you know that (x - b) is a factor, since there is no remainder. \n\nNow (x - b) can only be a factor of the polynomial if b is a factor of the constant at the end, in this case 5. So the possibilities to test are 5, -5, 1 and -1.\n\nTesting +/- 1 first as they are easiest\n\nPutting x = 1 gives 8 + 52 + 66 + 31 + 5 = 162.\nPutting x = -1 gives 8 - 52 + 66 - 31 + 5 = =4\nSo (x - 1) and (x + 1) aren't factors.\nPutting x = 5 clearly gives a positive value\nPutting x = -5 gives 0, so (x + 5) is factor.\n\nThe next step is long polynomial division.\n\n(x + 5)   )    8x^4 + 52x^3 + 66x^2 + 31x + 5\n\n8x^4 divided by x gives 8x^3 \n8x^3 * (x + 5) = 8x^4 + 40x^3\nSubtract this from the polynomial to find the remainder:\n12x^3 + 66x^2 + 31x + 5\n\nThe quotient so far is 8x^3\n\nRepeat this process to find the final quotient of:\n\n8x^3 + 12x^2 + 6x + 1\n\nIf you can recognise that this is of the form of a cubic:\n\na^3 + 3.a^2.b + 3.a.b^2 + b^3\nwhere a = 2x and b = 1, you have your solution.\n\nIf not, we again have to try to factorise by testing factors. Now it should be obvious that (x - b) can no longer be factor, since we know (x + 1) and (x - 1) can't be factors. We need to try something of the form (ax - b), by observing that x^3 has a coefficient of 8, so (2x + 1) is a reasonable factor to try, since 2 is a factor of 8 at the start of the polynomial and 1 is a factor of 1 at the end of the polynomial. You can test if it's factor, because if (2x + 1) is factor, then x = -1/2 must be solution to the polynomial being equal to 0, which it is.\n\nPerforming polynomial division again gives 4x^2 + 4x + 1, with the factors now at (x + 5)(2x + 1). From here you can use trinomial factorising techniques to get the final solution of (x + 5)(2x + 1)^3
824178
You typed exponents just fine!\n\nx^3 - x^2 -x + 7 divided by x+1 is a "long division" type problem rather than factoring because x+1 doesn't go evenly into the polynomial\n\nThe answer is x^2 -2x + 1 + 6/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
471239
Here's how to complete the square:\n\nI'll use the example of \n\n2x^2 - 4x -6 = 0\n\n1.  Divide all the terms by the coefficient of the x^2 term.\n\nThe coefficient of the x^2 term in your example is 2, so\n\n(2/2)x^2 - (4/2)x -(6/2) = 0\nx^2 - 2x -3 = 0\n\n2.  Bring the constant term to the right side of the equation.\n\nx^2 - 2x = 3\n\n(This is where you are at in your example)\n\n3.  Complete the square\n\nAny square of the form (x + a)^2 is x^2 + 2a + a^2.  Look at the x term in your example.  The coefficient is -2.  In your example, 2a is going to be -2.  So a=-1.  Right now, you have (on the right side)\n\nx^2 +2(-1)\n\nThe missing part of the square is (-1)^2, which is equal to 1.  You complete the square by adding that value to the right side.  You keep the euqation balanced by adding it to the left side:\n\nx^2 - 2x +1 = 3+1\n(x - 1)^2 = 4\n\n4.  Take the square root.  Remember that there are two possibilities:\n\nx - 1 = 2\nx - 1 = -2\n\n5.  Solve both equations:\n\nx = 3\nx = -1\n\n6.  Check both your answers.  Anybody in the world could make an error here.  The most important thing is to check both of your answers:\n\nx = 3\n\n2x^2 - 4x -6 = 0?\n2(9) -4(3) -6 = 18 -12 -6 =0\n\nx = -1\n\n2x^2 - 4x -6 = 0?\n2(1) -4(-1) -6 = 2 + 4 - 6 = 0
469289
3x^2 = 12\n\nFirst, divide by three.\n\nx^2 = 4\n\nTake the square root.\n\nx = +2, -2
572480
breaking it apart, we have:\n\n(x+1)^3=(x+1)(x+1)(x+1)=\n(x^2 + 2x + 1)(x+1) =\n(x^3 + 3x^2 + 3x + 1) SO\n6(x+1)^3 = 6(x^3 + 3x^2 + 3x + 1) = 6x^3 + 18x^2 + 18x + 6\n\n(x+1)^2 = (x+1)(x+1)=\nx^2 + 2x + 1   SO\n-3(x+1)^2 = -3(x^2 + 2x + 1)=\n-3x^2 - 6x - 3\n\n4(x + 1) = 4x + 4\nand -9 stays the same.\n\nSo putting everything together, you have\n(6x^3 + 18x^2 + 18x + 6) + (-3x^2 - 6x - 3) + (4x + 4) - 9 \ncombine like terms and you get:\n6x^3 + (18x^2 -3x^2) + (18x - 6x + 4x) + (6 - 3 + 4 - 9)\n6x^3 + 15x^2 +  16x - 2\n\nWHEW!  That took me back to college pre-calc!
660785
i don't fucking know.\n\nEDIT: waitaminnit i'm gonna give it a try!\n\nx^3-4x^2-7x+10 = 0\n\nfor x = -1\n-1 -1 +7+10 != 0\nso x != -1\n\nfor x = 1\n1-1-7+10 != 0\nso x != 1 either!! ...goshdarnit.\n\nx^3-4x^2-7x+10 = 0\nx(x^2-4x) - 7x = -10\nx^2-4x - 7 = -10/x (we divided both sides by x)\n..ok...i think i've lost the plot sorry.
798514
do you mean 6x([1/2]x^2)?\nif so 3x^3\nor 6x(1/(2x^2))?\nthen its 3x^-1
484892
(x+3)(2x+4) is incorrect.\n(x+3)(2x+4) = 2x^2 + 10x + 12\nSolution:\nMultiply the coefficient of the x^2 term with the constant.\n2(12) = 24\nYou are looking for factors of 24 that add the 11 (which is the coefficient of the x term).\n(3)(8) = 24\n3 + 8 = 11\nReplace 11x with 3x + 8x\n2x^2 + 11x + 12\n= 2x^2 + 3x + 8x + 12\nFactor by grouping\n(2x^2 + 3x) + (8x+12)\n= x(2x+3) + 4(2x+3)\n= (2x+3)(x+4)\nAnswer: (2x+3)(x+4)\nYou can easily check the answer by Foil.\n(2x+3)(x+4) = (2x)(x) + (2x)(4) + (3)(x) + (3)(4)\n= 2x^2 + 8x + 3x + 12\n= 2x^2 + 11x + 12
570705
Factoring out X gives:\nX(-X^2 + X + 6)=0\nUsing the quadratic formula to factor the terms inside the brackets gives:\n-X^2 + X + 6 = (X+2)(-X+3)\nSo we have:\nX(X+2)(-X+3)=0 and the zeros are X=0, X=-2, X=3
615827
alright... \n\nso y=2x²-6x+1?\n\nthere could be 2 things coming out of this:a parabola or an absolute value, but i think the parabola's what you need.\n\nthe equation for the parabola is y=a(x-h)+k\nyou need to do the method 'completing the square'\nbut first factor out the 2 from the first 2 terms so it becomes y=2(x²-3x+?)\nto make a perfect square, take half of the third term, square that to make the third term. so half of 3 is 3/2, squared is 9/4.\nbut you can't just put a 9/4 because it was never originally there. so you have to either subtract it back out or add it to both sides.\nso it becomes y=2(x²-3x+9/4-9/4)+1\nsee, you add the 9/4 and you subtract it. the +1 was the number you originally had. then you factor the first 3 terms inside the parentheses\nso it would become y=2(x-3/2)²+1\n                   y=a(x-h)+k\nnow, the axis of symmetry would be (-3/2,1)\n to answer the second question, i dont know what the equation is\n\nfor the third question you have to plug in the numbers\n(x+3)(x-5) I'm not sure what roots are because its different for me but I'm thinking its the answer\n"what makes this equal 0?" is what you have to ask. so if u want -3. you want positive 3, and vise-versa for 5\nthen you FOIL method it, which equals x²-2x-15