i'm doing a polynomial expansion of (2/x+3x)^6 and i really dont  know how to begin it\nwhat would the absolute coefficient of x in the first term be?\ni just cant seem to pull out my algebra skills today! gosh!
207830
Did you mean (x/2 + 3x)^6 or is it really (2/x + 3x)^6\n\nIf it's the first, the coefficient of x is 1/2.  If it's the second way, though, the coefficient is 2, but it's times x^(-1).
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You asked about the numerator before, didnt you get an answer?\nWith synthetic substitution, you should find that the roots of the numerator are -1, -1, -2, and 3.\nThis gives (x+1)(x+1)(x+2)(x-3)  see *.\n\nAre you sure you typed the denominator correctly?\nYou typed: \nx^6 - 6x^5 + 15x^4 - 30x^3 + 15x^2 - 6x + 1\n(by the way, you need to insert spaces into long polynomials like this one or Yahoo does not display them after 30 spaces.  You can place the mouse cursor over them and see the complete polynomial, but many people dont know that, including the previous answerer.)\nThis is not factorable (over the rational numbers)\n\nHowever, if one digit is changed, namely if -30x^3 was written \n -20x^3, then the denominator becomes a very common one (and one that you may notice just by looking at it if you stay at this factoring stuff for a long time):\nnamely (x-1)^6 !!\n\n\n*\n1st root (-1):\n\n-1 || ..1... 1 ... -7 .. -13 ... -6\n...............-1 .... 0 ..... 7 ..... 6\n————————————\n..........1....0.....-7.....-6... || 0\n\nso (x+1) is a root\n\n2nd root (-1):\n\n-1 || ...1....0.....-7.....-6\n................-1 .... 1 .... 6\n——————————\n..........1...-1 ... -6 .. || 0\n\nso (x+1) is a root again.\n\nx^2 - x - 6 remains\n= (x-3)(x+2)\n\nInteresting correlative problem:\nhttp://answers.yahoo.com/question/index;_ylt=Aqz3W3s3NPcz.j__CQYxK0bsy6IX?qid=20061127003230AA2oHuA
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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.
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start: (a^2+2a+1)(a+1)\n\nyou basically multiply all the terms\n\na^2(a)+a^2(1)+2a(a)+2a(1)+1(a)+1(1)\n\nwhich is a^3+a^2+2a^2+2a+a+1\n\nwhich is a^3+3a^2+3a+1\n\ntoo bad i've completely forgotten how to go the other way...
486916
Horners' method of synthetic division is used to find the roots of polynomials. \nwe have to know one of the roots to proceed with this. we can check if 1or -1 are roots of the polynomial using some easy tricks. \n1) if the sum of all the coefficients is equal to 0, then 1 is a root\n2) if the sum of coefficients of even terms is equal to the sum of the coefficients of the odd terms, then -1 is a root.\nnow, for the first equation, we already know that -2 is a root, because the question is to find the quotient when the polynomial is divided by x+2. \nin synthetic division, u first put down a 0 beneath the first coefficient, ie., 3. now add these two. the result is 3. \nnow multiply 3 with the root, ie., -2. the result is -6. write this beneath the second coefficient, ie., 2. the result is -4 after addition. \nrepeat the same steps until u reach the last coefficient.\nnow next we write 8 below -1. result is 7.\nnext we write 7*-2 ie., -14 below 5. the result is -9.\nlastly, we write -9*-2=18 below . the result is 12, but the result should be zero.\nso, the given equation is not divisible by x+2.\n\ncoming to the second equation, follow the same procedure as above and u will get the resultant coefficients at each stage to be 2,6,22,61 respectively. so it is not divisible by 3
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The zeros tell us that x-1, x-2 and x-4 are factors, so the polynomial is  \nf(x) = k*(x-1)(x-2)(x-4) where k is some constant.\n\nf(0) = k*(-1)(-2)(-4) so you just choose the value of k that makes this -16.\n\nIf 3 + i is a root then so is 3 - i, and multiplying\n(x - (3+i))(x - (3-i)) gives\n\nx^2 - 6x + 10, so this is your quadratic polynomial.\n\nObviously the linear factor is x - 4 since  4 is a root.  Leading coeff 1 means you don't have to bother to multiply these factors by some other constant coefficient.
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3(x^4+3x^2-11)\n\nseriously you need to pay attention in class instead of asking people. lol
19694
Sorry it's not easy to draw polynomial division here.\nAnyway: the first remainder (if I calculated without errors!) is -a^2-a^3\nThe remainder of the second division is a^3-2a-3.\nYou then have the equation:\n-a^2-a^3=a^3-2a-3\nthat is\n2a^3+a^2-2a-3=0\n\nYou can also write this as\n2a^3+a^2=2a+3\n\nNow you can find a graphical solution to the equation by searching the intersection between the cubic y=2t^3+t^2 and the linear y=2t+3, and considering the value of the ascissa of such intersection.\n\nOtherwise, any other method for solving cubic equations can work.
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alright, you need to understand this much more than you need the correct answer. You won't have Yahoo! answers to help you on any tests.\n\nYou need help factoring. A real-life tutor who can talk to you and show you in person is best, but I'm going to try to help you understand. \n\na polynomial is in the form ax^2 + bx + c. This is from the formula of (Z*x + q) * (R*x + p). Don't let all these letters scare you. They are just place holders for numbers. \n\nThe only thing that can get the letter A in Ax^2 +bx +c is the first two parts of the "terms" you need to multiply together. With letters a= z*r. In order to get C, you only need to worry about the last parts in the terms. Using letters c= q*p. B is what makes sure you know what two terms were multiplied together. B = z*p + q*r. In order to "factor" out a term from a polynomial you have to find the set of z, p, q, and r that works. I'll show you now with numbers.\n\nLet's say you have the equation 2x^2 + 8x +8. See, it's in the form of ax^2 +bx + c. a=2, b=8, c=8. NOw you have to figure out the facotrs of 2 and 8 (which are a and c)\nfor a=2 the numbers that are multiplied together to get two(the factors) are 2 and 1. Since you know that only the first two parts of the things you multiply together have to get you a, you can set up something that looks like this: \n(2x + ?) *(x + ?) \n\nNow you need to find what two numbers fit into your equation to get you c=8. The factors of 8 are 1*8 or 2*4. Let's try putting 1*8 into the equation.\n(2x+1)* (x+8) = 2x^2 +17x + 8. That's not right, so lets try to reverse them.\n(2x+8)* (x+1) = 2x^2 +10x + 8. That's not right either, so we know only the factors of 2*4 will work in this equation.\nLet's try them\n\n(2x+2)*(x+4) = 2x^2 +10x +8. It may seem like we've tried each of the factors, but we can still reverse 2*4.\n\n(2x+4) * (x+2) = 2x^2 +8x+8. That's right. Once you get more practice and understand this better, you can "see" certain factors that have to be included just because "factoring" will be automatic. But, just to make sure you should always check you answer like I did when I showed you the example.\n\nNow that you hopefully understand the hard part, perfect squares are easy. All you do is take  the square root of 'a' and take the square root of 'c' and put them into the equation. When it is a PERFECT SQUARE this works because of: ax^2 + bx +c = (z*x+ q)^2. ax^2 = (zx)*(zx) so a = z^2, and the square root of a = z. c = (q)*(q) so c = q^2, and the square root of c= q. I'll use numbers now.\n\n100x^2 + 240x + 144. a = 100, so the quare root of 100 = 10. z=10. Then the square root of 144 is 12, so q=12. That means y=(10x+12)^2. Now, to find b, all you have to do is foil it out.
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ill give it a shot - im not sure , anyways i got 2 answers..\n\n(1st)\n\n(3-x)x+2 = 2 (x^2-2)\n\n3x-x^2+6-2x = 2x^2-4\n\nx-x^2+6 = 2x^2-4\n\nx-1/x+6 = 2/x-4\n\nx-1+6x = 2-4x\n\nx+4x+6x = 2+1\n\n11x = 3\n\nx = 3/11\n\n\n(2nd possible answer)\n\n\n(3-x)x+2 = 2 (x^2-2)\n\n3x-x^2+6-2x = 2x^2-4\n\nx-x^2+6 = 2x^2-4\n\nx-1/x+6 = 2/x-4\n\nx-1+6/x = 2-4/x\n\n-5 = -2/x\n\n-5x = -2\n\nx = -2+5\n\nx = 3
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Monomial factor is 3x^4  The factorization is\n\n3x^4 * (2x^3 + 5x^2 - 3x - 3)\n\nI cannot see how 2x+1 is a factor of 2x^3 + 5x^2 - 3x - 3; polynomial division does not result in integral coefficient and also leaves a remainder.
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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
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I also have difficulty typing this thypes of math problems.  First, simplify the ax^2+bx+c parts in the first half to (x+c)(x+c) looking things first.  So...\n\n[(x-3)(x-1)]} is the first part.  With the bottom of the second half, divide by 3, so that you get 3(x+1).  Now line everything up like so...\n\n(x+1)(x+3)(x-3)\n_____________\n3(x-3)(x-1)(x+1)\n\nNow do the cancellations and you get the right answer...\n(x+3)\n____\n3\n\nSorry for not working to convention, but it was easier to type it that way.
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Do you mean real AND imaginary? Or just all imaginary?  Well, either way, since we \nhave a polynomial with four factors, we will factor by grouping to simplify.\n\nf(x)=(x^3)-(9x^2)+26x-24\n\nf(x)=[(x^3)-(9x^2)+26x]-24\n\nf(x)=x[(x^2)-9x-26)]-24\n\nf(x)=(x-24)[(x^2)-9x+26]\n\nWe have one real solution. Our linear factor is (x-24), so our Real solution is x=24.\n\nNow we need to use the quadratic formula:\n\n{-b(+or-)[The Square Root of:(b-4ac)]}ALL over 2a.\n\n[(x^2)-9x+26] a=1, b=-9, c=26\n\n-(-9) ± √(-9-(4*1*26))ALL divided by 2\n\n9 ± √(-9-104)ALL divided by 2\n\n9 ± √(-113)ALL divided by 2\n\nSo our two imaginary solutions are:\n\n [9+√(-113)]/2] and [9-√(-113)]/2]\n\nRecall that i=√(-1)\n\nSo, our first solution [9+√(-113)]/2]=9+i√(113)or 9+√(113)i\n\nand our second solution [9-√(-113)]/2]=9-i√(113)or 9-√(113)i\n\nTherefore, All Solutions, Real and Imaginary, for \n\nf(x)=(x^3)-(9x^2)+26x-24 are:\n\nx=24, x=9+√(113)i, and x=9-√(113)i\n\n:)
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When you have a polynomial, the coefficients are integers.  For example:\n\nax^2 + bx + c = 0\n\nThis is the general formula for a second degree polynomial (x is to the 2, hence second degree).  In this case, a, b, and c are the integers.\n\nAlso, if you add other terms, such as x^3, they would also be multiplied by integers.
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(x²+4x+4) / (2x-3) / (x²-4) / (2x²-7x+6)\n\nYou have 4 terms and three divisions, so it's a little unclear if you mean (a/b) / (c/d) or what.  I will try to help to the extent I can.\n\nLet's assume you mean:\n\n[ (x²+4x+4) / (2x-3) ] / [ (x²-4) / (2x²-7x+6) ]\n\nFactor your x² equations to see if you can cancel terms:\n\nx²+4x+4 = (x+2)(x+2)\nx²-4 = (x+2)(x-2)\n2x²-7x+6 = (2x-3)(x-2)\n\nWoohoo - things are looking good!  Now we can rewrite our original problem using these factors:\n\n[ (x²+4x+4) / (2x-3) ] / [ (x²-4) / (2x²-7x+6) ] = \n\n[ (x+2)(x+2) / (2x-3) ] / [ (x+2)(x-2) / (2x-3)(x-2) ] =\n \n[ (x+2)(x+2) / (2x-3) ] * [ (2x-3)(x-2) / (x+2)(x-2) ] =\n\nnotice we can cancel out three terms: (2x-3), (x-2) and (x+2)\n\n[ (x+2)(x+2) / (2x-3) ] * [ (2x-3)(x-2) / (x+2)(x-2) ] =\n\n(x+2)\n\nHope that helped!\n\n~ ♥ ~
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(x+7)(x+3)\n(x-11)(x+2)\nforget this one.  its AC factoring though\nthis ones AC too