Can anyone help me factor this rational function? I'm stucked! Thnks!?
818643
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
752879
ok re arranged the second question so that the 6 is factored out so its x= such and such then sub the x equals such and such into the first equation and solve then do the same thing to Y in the second equation
865391
I think there is a sign error: the cubic term is supposed to be -2x^3, not 2x^3. Then the other roots are 3 and -1 and the factors are\n\n(x-8i)(x+8i)(x-3)(x+1)
190708
you've asked a lot of questions regarding rational and irrational. Here's a site that may be of help:\n\nhttp://www.purplemath.com/modules/numtypes.htm
909578
Since the coefficient of x^4 is one, the possible rational zeros are just the factors of 8.\nList all the factors of 8, both the negative and positive.\nAnswer: -1, +1, -2, +2, -4, +4, -8, +8
780960
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.
930308
x=[2+/-rt(4+60)]/2\n=[2+/-8]/2\n=-3 or 5\nso the factors are\n(x+3)(x-5)
598723
possible rational factors are p/q where, p is factors of the constant and q is factors of the leading coefficient.\np = 15 and q = 2\nfactors of p: 1, 3, 5, 15\nfactors of q: 1,2\nPossible p/q = (plus or minus) 1, 3, 5, 15, 1/2, 3/2, 5/2, and 15/2\nKeep picking one of these values and test it until you get a remainder of 0.\n1 works\n1 2 3 -11 -9 15\n ___ 2  5  -6 -15\n__2 5 -6 -15 0\nThus,\n2x^4 + 3x^3 - 11x^2 - 9x + 15 \n= (x-1)(2x^3 + 5x^2 - 6x -15)\n2x^3 + 5x^2 - 6x -15 can be factored by grouping.\n(2x^3 + 5x^2) + (- 6x -15)\n= x^2(2x+5) -3(2x+5)\n= (x^2-3)(2x+5)\nAnswer: (x-1)(x^2-3)(2x+5)
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
787164
is there a specific area that you need help with in rational numbers?  I guess you can use a site that i know of.  Here: \nwww.algebra-online.com\n^ after that put a slash and write: rational-numbers-1.htm\nthat might help you (:
859744
(x + 4y)(x - 4y)\nI answered 2 of your questions. Do the rest of the questions yourself.
894442
f(x)=2x^3-3x-1\nf(-1)=-2+3-1=0\nx+1 is a factor\ndividing by x+1\n2  0  -3  -1\n0 -2   2   1\n-------------\n2x^2-2x-1\nso the factors are\n(x+1)(2x^2-2x-1)
498372
I think u are talkin abt this method....\n\n\nif u hav a cubic like  x^3 + 2x^2 - 2x - 1 = 0\n\nto find out whether (x-1) is a factor....\n\nfirst write the coeff of the eq in this way n write the factor ( here it iz 1 since x-1=0 or x=1) on the left hand side Like this\n\n{one}  1     2     -2     -1\n--------  0     1      3      1\n_______________________________\n{sum}  1     3      1      0\n\n\nHere what i hav done is write the 0 under the coeff of x^3\nnow add that coeff n the zero\nu get 1...\nnow multiply this one wid the number written on the left hand side(here it is also one)\nwite what u get under the coeff of x^2....\nas above add the new no and the coeff of x^2\nin a similar way multiply one and the sum u got n write it under the coeff of x...\nadd the no with coeff of x n then write it below....\nfinally u multiply the sum wid one and write it under the constant term.... n then add...\nif u get 0 as the sum then whatever no u hav taken in da LHS is on of the root.... here we get 0 in the last case... so (x-1) is a fctor.....\n\nnow this eq can b written as \n\n    (x-1) * (1*x^2 + 3 * x + 1) = 0\nhere 1 , 3 and 1 of the quadratic are the sum u got in the above method......
842037
We need to factor the equation down. This involves finding the roots of the equation. This will look like:\n\n(ax+b)(cx+d) = 12x^2-31x+20 = 0\n\nBecause the final term is positive and the middle term is negative, my first hunch is to have b = -4 and d = -5. By doing so, the equation looks like:\n\n(ax-4)(cx-5) = 12x^2-31x+20 = 0\n\nThe factors needed for 12x^2 could be 2 and 6 or 3 and 4. We should test each combination to see that they work out. The factoring ends up looking like this:\n\n(3x-4)(4x-5) = 0\n\nNow all that remains is to solve for x in each set of parentheses. If the first term is zero, then\n\n3x-4 = 0\nx = 4/3\n\nIf the second is zero, then\n\n4x-5 = 0\nx = 5/4\n\nThus, x = 5/4, 4/3
195306
I haven't done factoring for a while but I'm pretty sure this is right.\n\nSo you start with 125q²-n³q²\nIf you take out both the q² 's you have q^4 \n(q x q x q x q)\nSo you are left with q^4(125-n³)\nThe cubed root of 125 is 5\n(5 x 5 x 5)\nand the cubed root of n³ is n\n(n x n x n)\nSo your answer is q^4(5-n)\n\nI think that's right.
547775
f(x)= x^3+x^2-22x-40\nf(x)= (x+4)(ax^2 + bx + c)\n\nComparing coefficients,\nx^3: 1=a\nconstant: -40=4c => c=-10\nx^2: 1=b+4a => 1=b+4 => b=-3\n\nf(x) = (x+4)(x^2-3x-10) = (x+4)(x+2)(x-5)\n\nThe other factors are (x+2) and (x-5).
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.