How do I mulitply Polynomials? in this problem. . .?
921411
(x+2)(x-5)=(x-1)(x-3)\nx^2-3x-10=x^2-4x+3\n-3x-10=-4x+3 [x^2's cancel]\n+3x       +3x\n-10=-x+3\n-3        -3\n-13=-x\nx=13
863763
D ( 2 or 0)\n\nThis is because the coefficients of the given cubic polynomial are real numbers and so any real zero needs be accompanied by its complex conjugate and so as stated  earlier MUST come in pairs so either there are 2 or there cannot be any.
917477
where else?\n\nhttp://www.factoring-polynomials.com/
703053
so the Q is basically asking for two binomials which, if multiplied will end up as a polinomial...\nAs a matter of fact any two binomials can form a polynomial.\n(2X+3) (x-7)...this is an e.g.
461122
Since f(x) is a polynomial of degree two in x,\nLet f(x) = ax^2 + bx + c, where a,b,c are constants.\n\nLet's find f(x) first.\n\nf(0) = 1 implies, c = 1.\n\nf(1) = 3 implies, a + b + c = 3\nor  a + b = 2 , or b = 2 - a  ...(1)\n\nf(-1) = 5 implies, a - b + c = 5\nor a - b = 4 ...(2)\n\nSubstituiting (1) into (2), we get,\n\na - [ 2 - a  ] = 4\n2a = 6 or a = 3.\nThus, b = 2 - a = 2 - 3 = -1.\n\nHence, f(x) = 3x^2 - x + 1\n\nNote that by long division or by the identity below,\n\n3x^2 - x + 1 = 3(x + 1)^2 - 7x - 2\n             \nWe get,\n\n[ 3x^2 - x + 1 ] /[1 + x]^2 = 3 - [7x + 2]/ [1 + x]^2   \n\nAnd that together with below,\n\n[7x + 2]/ [1 + x]^2  = 7 /(x + 1) - 5 /(x+1)^2\n\nGives,\n\n[ 3x^2 - x + 1 ] /[1 + x]^2 = 3 - 7 /(x + 1) + 5 /(x+1)^2\n  \nNow, we have an "integratable form".\n\n∫ [f(x) / (1 + x)^2] dx\n\n= ∫ [3x^2 - x + 1] / (1 + x)^2] dx\n\n= ∫ [ 3 - 7 /(x + 1) + 5 /(x+1)^2 ] dx\n \n= 3x - 7 log(x + 1) - (5/[x + 1]) + C\n\nwhere C is any arbitary constant.\n\nNote: Sometimes log(x + 1) can also be written as ln (x + 1)\n\nRemark. It is generally accepted that log and ln are both in base "e", the Euler number. If its in a different base, then usually its said or stated. For example logarithm of x in base 2 can be written as log_2(x).\nJust symbols :)\n\nRegards
825795
OK,as the first one is already done,I am doing the rest for you\n4(2x-6)=x+2 =>8x-24=x+2  =>8x-x=2+24  =>7x=26  =>x=26/3=8 2/3  (3) 2(x+5)+1=3(x-1)+4   =>2x+10+1=3x-3+4 =>2x-3x=-3+4-10-1  => -x=-4  =>x=1  (5)x+6=4(4x-1)   =>x+6=16x-4  =>x-16x=-4-6  => -15x=-10  =>x=-10/-15=2/3 [after reducing]
241036
I'm assuming you're looking for information on factoring polynomials since it's a College Algebra courese.  \nThis website is very helpful, with the steps for factoring laid out, and it also provides websites you can use if you need more information.  Good luck!\n\nhttp://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut7_factor.htm
208169
Factoring, I guess, but it really depends on the type of polynomial question. Can you give a sample?
76424
You can reduce you work by knowing that if your polynomial looks like  Ax^n + ... + B that if you have a factor that looks like (ax + b) that a has to be a divisor of A and b has to be divisor of B. That means that the only possibilitys in your example are\n(x + 1), (x - 1), (-x + 1) and (-x -1)\nSince (-x - 1) = -1(x+1) and (-x + 1) = -(x - 1)\nyou only have to look at two cases (x+1) and (X-1)\n\nTo find out if either of the two is a factor, you can do long division on the polynomial just like you would with regular numbers. If you don't know how to do that look at the link below for an example.\nI can't show you here because it won't format correctly.
503153
(x+7)(x+3)\n(x-11)(x+2)\nforget this one.  its AC factoring though\nthis ones AC too
928663
Are you asking for help or an answer?\n\nTake any 2 equations and multiply all of the terms or one or both by numbers that will make one of the terms in each equation have the same coefficient. Now subtract one from the other and the term will disappear. Now take the resulting equation and do the same with it against the third. Keep doing this until you are down to one term and solve for the variable. Now substitute the variable into the equation and solve for the other variables.
743452
Assuming that x^4 is not in the denominator:\n\n(x-2/x^2) simplifies to  (x^-1) - (2x^-2)\n\nTherefore, the problem becomes:\n\nint[ x^-1  -  2x^-2  + x^4 ]\n\nNow you can use the power rule to integrate each term:\n\nln(x) + 2x^-1 + (x^5)/5  +  C
25290
If you're looking for a simple polynomial expansion, you follow this simple rule: (A+B)(C+D)=AC+AD+BC+BD. In your case, 8x*3x+8x*(-1)+5*3x+5*-1. Solving, this becomes: 24x^2-8x+15x-5\nGrouping like terms: 24x^2 + 7x - 5 .\n\nIf you're solving for x, set each term = 0. In other words: 8x+5=0 and 3x-1 = 0. Solving, x=-5/8 and x=1/3.
221826
1) \nxy - x + 4y - 4\nx(y-1) + 4(y-1)\n(x+4)(y-1)\n\n2)note: x^3 means x cubed; x^2 means x squared\nx^3 -6x^2 + 2x - 12\nx^2(x-6) + 2(x-6)\n(x^2 + 2)(x-6)\n\n3)\n6tx + r - 3t - 2xr\n3t(2x - 1) - r(2x - 1)\n(3t - r)(2x - 1)\n\n4)Are you sure it wasn't -x^2 - 2x - xy - 2y?\n-x^2 - 2x - xy - 2y\n-x(x + 2) - y(x + 2)\n(-x - y)(x + 2)\n\nTo "factor by grouping," you must split the original polynomial that you are given into two different binomials by "grouping" them together.  This is done with perenthesis in the problems above.  Then, you factor out the "greatest common factor" from each of the binomials.  After factoring, the two quantities still within the perenthesis should be the same.  You could call these quantities the "common binomial."  Lastly, you factor out the common binomial to get the final answer.
364430
Did you mean to have the yx^2-30x+8 as your polynomial (a polynomial in two variables?)\nFor y=7   or 7x^2-30x+8\n\n7x^2-30x+8 = 7x^2-28x-2x+8\n\n7x^2-28x-2x+8= 7x(x-4)-2(x-4)\n\n7x(x-4)-2(x-4)= (7x-2)(x-4)
1157065
(x+5)(2x^2+4x+2)\nx(2x^2+4x+2)+5(2x^2+4x+2)\n2x^3+4x^2+2x+10x^2+20x+10\n2x^3+14x^2+22x+10
157895
Not quite sure what you mean, but I think this is what you're asking for:\n\nif you have y = x^3 + 2x^2 - x + 5\n\nto differentiate that, let's rewrite the equation like this: \n\ny = x^3 + 2x^2 - x^1 + 5x^0\n\nNow, for each item in the equation, multiply that item by its exponent, and subtract 1 from the exponent, giving you:\n\ny' = 3*(x^2) + 2*(2x^1) - 1*(x^0) + 0*(5x^-1)\n\nwhich simplifies to:\n\ny' = 3x^2 + 4x - 1\n\nAnd there you have the derivative of a simple polynomial equation.