this expression is associated with polynomials
119382
3(16 y_squared - 24xy + 9 x_squared)
210673
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"
398036
You have to find the two polynomials that when multiplied by each other will equal the trinomial.\n\nx2 + 8x - 9\n\n(x + 9) (x - 1)\n\nx*x + -1x + 9x -9\n\nx2 + 8x - 9
397013
Look in your text book under "factoring polynomials"... or you could try this web site that explains this fairly thoroughly...\n\nhttp://faculty.ed.umuc.edu/~swalsh/Math%20Articles/Factoring.html
249158
I'm not sure if this is the whole question, but I will interpret it as if it ended with a ) in place of the [exp...\n\n9(x^3)N - 12(x^4)(N^2) - 9(x^2)N
469984
Coefficients are:  1  1  3
433926
Let g(x) = f(x) - f'(x). Then the problem can be restated as saying that g(x) has at least one real root. If f(x) is a polynomial of degree n, where n≠0, then f'(x) is a polynomial of degree n-1, and so the difference of the two polynomials will be a polynomial of degree n. We consider two cases:\n\nSuppose n is odd. Then g(x) has odd degree, and all polynomials of odd degree have a real root, so we are done.\n\nSuppose n is even. Then f(x) has either an absolute minimum or an absolute maximum. Suppose without loss of generality that it has an absolute maximum at c (the arguments for polynomials with an absolute minimum are essentially the same). Then at that point f'(c) = 0, so g(c) = f(c). Since f(x) has a real root, f(c) ≥ 0. If f(c) = 0 then g(c) = 0 and we are done. Otherwise, consider that the x^n term of f(x) dominates the terms of f'(x), so [x→∞]lim g(x) = -∞. Thus, g(x) is negative at at least one point, and since g(x) is everywhere continuous, then by the intermediate value theorem it must have at least one zero. Q.E.D.\n\nNote that in the special case where n=0, f(x) is a constant function, and thus does not have a root, so it does not meet the conditions of the theorem. Finally, in the case where f(x) = 0 (which is usually considered to be a polynomial of indeterminate degree, rather than degree zero, primarily to avoid exceptions to the fundamental theorem of algebra), f(x) is everywhere equal to f'(x).
384935
[(x^3)/(2 - x)]^4/3 \ndy/dx = (4/3)[(x^3)/(2 - x)]^1/3[3x^2(2 - x) + (x^3)]/(2-x)^2\ndy/dx = (4)[(x^3)/(2 - x)]^1/3[6x^2 - 3x^3 + (x^3)]/(3(2-x)^2)\ndy/dx = 8x^3(3 - x)/(3(2-x)^7/3)
14981
u=(1/3)x^2  and v=x-2x^2\n(u^3)v = ((1/3)x^2)*(x-2x^2)\n(u^3)v = ((1/9)x^6)*(x-2x^2)\n(u^3)v = ((1/9)x^7)-((2/9)x^8)\n\n\nNote: the notation used for u could also be interpreted as u=1/(3x^2) - if this is the case, use the same method substituting the correct equation for u (or post the question again with parenthesis).  Similarly, u^3v could mean either (u^3)*v or u^(3v), but again, the same method of substitution could be used to express the second expression in terms of x.
81235
If y is a root of this and b is in F_p, then (y+b)^p -(y+b)=y^p +b^p -y-b=a, so y+b is another root. Hence the complete list of roots is y, y+1, ....y+p-1.\n\nThus we have two cases, the first where the polynomial splits in F_p (a=0) and the other where we get a degree p extension, so F_(p^2).\n\nNotice that we are using the algebra student's dream (x+y)^p=x^p +y^p which is justified because we are working over a field of characteristic p.
61042
Megami of Nekro is mostly correct, but I will try to clarify the situation a bit. \n\n1. factorize, or factor\n\nIf you have an algebraic expression that can be described as the product of some simpler expressions, to rewrite the expression as this product is to factorize it. \n\nSuppose you want to factorize the polynomial x^2 -3*x. In the case of polynomials of some degree, we regard polynomials of lower degree as simpler. In this case, the polynomial x^2-3*x is of degree 2, because the biggest power of x is 2 (the x^2 term). The lower degree polynomials in this case are of degree 0, called numbers, and of degree 1, called monomials. We can write the polynomial x^2-3*x as the product of lower order polynomials in the following way:\n\nx^2-3*x=x(x-3)\n\nThe two monomials x and (x-3) are factors of the polynomial x^2-3*x, and by expressing the polynomial as a product of two lower order polynomials, we have "factorized it" or "factored it". \n\n\nHere is another one:\n\nx^4+2x^2+1 is factorized as \n(x^2+1)(x^2+1)\n\nWith grade 8 mathematics, you cannot go any farther than this. We replaced a 4th degree polynomial with the product of its two factors, two 2nd degree polynomials (remember, we have decided that lower degree polynomials are simpler)\n\n\nAside: It is also interesting to note that one can factor whole numbers sometimes into a product of smaller whole numbers, which we regard as simpler. Therefore, \n\n10=5*2\n\nand 24=2*2*2*3\n\nThis is also called factoring. Numbers that can be expressed this way are called composite numbers. Numbers that cannot be expressed this way are called prime numbers (the real definition is slightly more complicated: you can look it up).\n\n2. Expand\n\nTo expand an expression is usually the opposite of factorizing. You multiply all the parts of the expression, at least as much as you can, using the distributive law, etc. \n\nIn the case of the expression x(x-3), we multiply our two monomials and return to the polynomial x^2-3*x:\n\nx(x-3)=x^2-3*x\n\nHere is another example. To expand\n\nx*(y-2)*(z+5)*y\n\nwe just multiply these factors together, \n\n x*(y-2)*(z+5)*y=\n(xy-2x)*(zy+5y)=\nxzy^2+5xy^2-2xzy-10xy\n\n\n\n3. Simplify\n\nThere is some ambiguity about what it means to simplify an expression. Computer algebra programs that simplify expressions sometimes use one definition of the simplest expression, and sometimes another. It all really depends on what shows up the features you are more interested in at the time. \n\nUsually to simplify an expression you have to do things like:\n\na. add up any like terms\n\nTo simplify  5x + 2y+ 2x+3+4, we add the terms with just numbers in them, and add the terms with just xs in them:\n\n5x + 2y+ 2x+3+4=7x+2y+7\n\nb. Multiply any numbers that can be multiplied\n\n5*2*x+2+y simplified is usually 10x+y+2\n\nc. Usually, when one simplifies an expression, the big powers are presented first, and the small powers follow (or sometimes vice versa).\n\nx+x^3+x^2 can be simplified as either \n\nx^3+x^2+x or x+x^2+x^3, depending on taste, the needs of the moment, etc.\n\n\nd. Usually when simplifying, any numbers or polynomials in ratios that can be divided will be\n\nSo 10/6 is often simplified as 5/3\n\nAnd (x^2-3*x)/x is often simplified as x-3\n\nHopefully that should get you started.
308954
[-b +/-  sqrt(b^2-4ac)]/(2a)
199400
So, for a) & b) the two things that you should notice are:\n1 - The exponents inside the second polynomial alternate, meaning that as the powers of the x variable decrease, the powers of the y variable increase, both by increments of one.\n2 - The answer will only contain the variables x and y raised to the same power, which is one more than the leading exponent on the x variable.  \n\nSince (x + y)(x^2 - xy + y^2) = x^3 + y^3\n and  (x + y)(x^4 - x^3y + x^2y^2 - xy^3 + y^4) = x^5 + y^5,\nthen for any given polynomial the pattern is written out...\n(x + y)(y^2n - x^(2n-1)y + x^(2n-2)y^2 + ... + x^2y^(2n-2) - xy^(2n-1) + y^2n = x^(2n+1) + y^(2n+1)
544956
(A^2+2ab+b^2)/(a+b) + (a^2-b^2)/(a-b) + (a-b)/(a+b)/(a^2-b^2)/a+b)\n(a+b)^2/(a+b)+(a+b)(a-b)/(a-b)+ (a-b)/(a+b)*(a+b)/(a^2-b^2)\na+b+a+b+(a-b)/((a+b)(a-b))\n2a+2b+1/(a-b)\n\nif I got all the ( ) in the right places.
333751
[(2x^3+2x^2+3x+3)/(x^2-9)]*\n[(x^2+2x-15)/(4x^2+6)]\n\nFactor each expression individually.\n\n(2x^3+2x^2+3x+3) = (x+1)(2x^2+3)\n\n(x^2-9) = (x+3)(x-3)\n\n(x^2+2x-15) = (x+5)(x-3)\n\n(4x^2+6) = 2(2x^2+3)\n\nSubstituting these in, our expression becomes\n{[(x+1)(2x^2+3)]/[(x+3)(x-3)]}\n*{[(x+5)(x-3)]/[2(2x^2+3)]}\n\nCombining these into one larger expression, we get\n[(x+1)(2x^2+3)(x+5)(x-3)]/\n[2(x+3)(x-3)(2x^2+3)]\n\nCancel the quantities common to the numerator and denominator and we get our final answer\n[(x+1)(x+5)]/[2(x+3)]\n\nGood luck.
583346
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^_^
529747
(x+3)(x^2 + x + 2)\n= x(x^2 + x + 2) + 3(x^2 + x + 2)\n= x^3 + x^2 + 2x + 3x^2 + 3x + 6\n= x^3 + 4x^2 + 5x + 6