factoring polynomials and things...?
925669
You need to use the quadratic equation...\n\nhttp://mathworld.wolfram.com/QuadraticEquation.html\n\nOR\n\nx (x - 5) = 0
703982
1.\n3x + 6/((x + 1)(x + 2)) Factor 3 out of the top.\n[3(x + 2)]/[(x + 1)(x + 2)] x+2's cancel\n3/(x + 1) <-- This is simplified.\n\nI don't have the energy to do the other ones right now.  :/
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.
760715
If it is divisible by x - 3, then (x - 3) is a factor of that polynomial, and if x - 3 = 0, x = 3.  \n\nIf we substitute x = 3 to the polynomial, then it must yield 0 (since 3 is a root).\nx³ + 3x² + kx - 24\n\nThus,\n(3)³ + 3(3)² + k(3) - 24 = 0\n\nThen,\n27 + 27 + 3k - 24 = 0\n\nTranspose,\n3k = -30\n\nThus,\nk = -10\n\nThe polynomial then becomes\nx³ + 3x² - 10x - 24 = 0\n\nWe then divide it by x - 3, to factor it out.  The easiest method goes like this\nx³ - 3x² + 6x² -18x + 8x - 24 = 0\n\nNow, we group\n(x³ - 3x²) + (6x² - 18x) + (8x - 24) = 0\n\nWe factor out from each group\nx²(x - 3) + 6x(x - 3) + 8(x - 3) = 0\n\nWe then factor out x - 3\n(x - 3)(x² + 6x + 8) = 0\n\nWe can factor the quadratic trinomial\n(x - 3)(x + 2)(x + 4) = 0\n\nTherefore, your polynomial is also divisible by (x + 2) and (x + 4).\n\n^_^
866735
I believe you use the remainder theorem for this, which states that the remainder when a polynomial f(x) is divided by (x-k) is\nf(k)\n\nThus the remainder would be f(-1) = 11
91753
factorize\nx^2 + x = x(x+1)\n\nfactorize\nx^2 - 1 = .... (you should know)\n\n2 = 2 (!!!)\nTake the lcm of the three denominators (each different factor (=parenthesys) once and with the highest power)\n\nIt is 2x(x+1)(...)   (fill the dots)\n\n.....\n\nnow if you can't go on, you need to open your book again and study, or pay more attention at school at your math classes, and ask immediately for further explanation to your teacher when you feel you've not understood.\n\nGood work and good study
967957
There really isn't any way to do this other than guess and check. If you find that you substitute some value in and find f(a) = 0, then (x-a) will be a factor. You can repeat that a few times until you have found some roots, and then do a long division to something which you *can* solve by hand (a quadratic or linear).\nNote also that you only need to check things which are factors of the constant term, and there aren't too many.\nIn the first case, you will find that substituting in -2, -1 and 3 will all give 0, but you may have trouble finding the last root. So you will have to do some long divisions (divide by (x+2)(x+1)(x-3)), and you will end up finding that the -1 was a repeated root - so the answer is (x+2)(x+1)^2(x-3).\n\nSame thing for the second: you will find -1 as one root. You can then do a long division and find -1 is repeated, and again, and again, and again. The actual factorisation is (x+1)^5 - you may have noticed this if you knew much about pascals triangle and recognised the coefficients are an entire row - 1,5,10,10,5,1.
865971
p(x)=-3x^2 (x-1) (x+1/3) (x+2)^3\n\n1) what are the roots and multiplicities? \nroots - multiplicity\n0 .............2\n1..............1\n-1/3 .........1\n-2 ............3\nbasically the multiplicity of a root A is the exponent of the factor (x-A)\n.
968026
I got   3(x² - 5x + 4) or 3( x - 4 )( x - 1 )\n\nLet me show you my work...\n\n3( x - 2 )² - 3(x - 2) - 6\n\nSimplify\n\n3( x -2 )( x - 2 ) - 3( x- 2 ) - 6\n\nDistribute\n\n3( x² - 4x + 4) - 3x + 6 - 6\n\nDistribute and Simplify\n\n3x² - 12x + 12 - 3x\n\nSimplify\n\n3x² - 15x + 12\n\nYou can now simplify by factoring out 3\n\n3(x² - 5x + 4)\nor\n3( x - 4 )( x - 1 )
390615
For #1, try fully factoring the numerator and the denominator. You will find that some of the terms cancel. Start by factoring x from each one. Your resulting term in the numerator is a difference of squares. Your resulting term in the denominator is a factorable quadratic.\n\nFor #2 the expression will be undefined for any value that makes the denominator equal to zero. Again factoring so you can solve for the zeroes is the way to go.\n\nI don't like to work out problems for people, so I hope this is helpful. Likely enough someone else will give you the answers, but I wanted to make sure you knew the process.
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.
1053403
Write it as:\n2x^2-4x+3x-6\n=2x[x-2]+3[x-2]\n=[2x+3][x-2]\nSimple middle-term factorisation!!!
283601
1) First we move all the factors with x to one side, and the ones without x to the other.  Result will be:\n\n4x/3 - 2x + x = 3/2 + 1/5\n\n2)  Then, we factor out an x from the left side and get:\n\nx(4/3 - 2 + 1) = 3/2 + 1/5\n\n3)  Calculate the actual numbers on both sides\n\nx(4/3 - 6/3 + 3/3) = 15/10 + 2/10\nx(1/3) = 17/10\n\n4)  Divide both sides by 1/3 and get a final result of...\n\nx = (17/10)/(1/3)\nx = (10/17)*(3/1)\nx = 30/17\n\n5)  Comform that I PWN n00bs.  And... I'm pretty sure I pwn n00bs.  Therefor, we are done!  ^-^
208014
Factorising gives\n\n(x-4)(x+1) = 0\n\nThe solutions are therefore x = +4 and -1
890098
I enjoy algebra, and this is easily my favorite type of problem.  So I hope you, too, one day enjoy them for their inner beauty.\n\nFor now, let's tackle this "factoring problem."  We may wish to use the method of "synthetic division" which is arguably easier than "long division of polynomials."\n\nThink back to grade school where you learned that 6 = 2 * 3.  This simple statement enabled you to 'deconstruct' the number 6 as either "two times three" or "three times two".  Even if you forgot one of the 'factors' you could conjure it up again quickly by asking yourself, "What is six divided by two?" for example.\n\nIn problem (a), you are essentially asking, "What are the coefficients of P(x) if the only factor we know is (x-1)?"\n\nLet's set up synthetic division (see link below), by placing all the coefficients of P(x), and using 1 (the zero implied by (x-1)) as the divisor:\n\n1 ||  a  b  c  d\n___________\n.....a\n\n1 ||  a  b  c  d\n_____a_____\n.....a (a+b)\n\n1 ||  a  b        c       d\n_____a___(a+b)____\n.....a (a+b) (a+b+c)\n\n1 ||  a  b        c                d\n_____a___(a+b)___(a+b+c)_\n.....a (a+b) (a+b+c) (a+b+c+d)\n\nNote that this last term must equal 0 since it is assumed that (x-1) is a factor ... thus, the "remainder" term must equal zero!\n\nThere's the result you were after:  a+b+c+d = 0\n\nHope this helps you tackle the others!
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
631080
You can noy factor it out... it makes it easier to figure out when you put it into lowest terms... so it would be \n\n   x^2+7x-25..... so you cant factor it.