Math: Division of Polynomials:x^5+x^4-2x^3+2x^2-3x-7divided by x-1Answer=x^4+2x^3+2x-1-(8/x-1) Show how?
443017
Set it up like a long division problem:\n\nx - 1 ) x^5 + x^4 - 2x^3 + 2x^2 - 3x - 7\n\nNow focus on the x terms.  x is on the outside, and x^5 is on the inside.  Just like long division with numbers, how many times does x go into x^5?  x^4 times.  That goes on top, and then you multiply that by x - 1 and subtract, again like normal long division.\n\nx^5 + x^4 - 2x^3 + 2x^2 - 3x - 7\n-(x^5 - x^4)\n--------------------\n2x^4 - 2x^3 + 2x^2 - 3x - 7\n\n\nRepeat process...x goes into 2x^4 2x^3 times:\n\n2x^4 - 2x^3 + 2x^2 - 3x - 7\n-(2x^4 - 2x^3)\n--------------------\n2x^2 - 3x - 7\n\nx goes into 2x^2 2x times:\n2x^2 - 3x - 7\n-(2x^2 - 2x)\n--------------------\n-x - 7\n\nFinally, x goes into -x -1 time:\n-x - 7\n-(-x + 1)\n--------------------\n-8\n\nSo you have x^4 + 2x^3 + 2x - 1 with a -8 remainder, which you can write as -8/(x - 1).
985294
you just keep on deriving.\n\nI find it easier to split things into exponents.\n\nI am assuming your question goes\n\n(1/(2x))  - 4 = g(x)\nor\n\n(1/2)*x^-1     - 4 = g(x)\n\ndifferentiate once\n\ng'(x) = -(1/2) * x^-2 \ng''(x) = -2 * (-1/2) * x^-3  = x^-3\ng'''(x) = -3 * x^-4  =  -3/(x^-4)  **ans**\n\nPlug 3 into this... bah\n\nFor reference please use parenthesis to make your problem clearer for us. I just solved the whole thing using the wrong g(x).\n\n\n===!using the correct g(x) which should be g(x) =1 / (2x -4)===\n\ng(x) = 1 / (2x -4) = (2x - 4)^-1  don't forget the chain rule\n\ng'(x) = -1*2*(2x - 4)^-2 = -2 * (2x -4)^-2    chain rule again\n\ng''(x) = -2*-2*2*(2x -4)^-3 = 8*(2x -4)^-3    chain rule again\n\ng'''(x) = -3*2*8(2x - 4)^-4 = -48 * (2x -4)^-4\n\ng'''(x) = -48/[(2x -4)^4]\ng'''(3) = -48 * (6 -4)^-4 = -48/ (2^4) = -48/16   = ***real ans***\n\ng'''(3) = -3
414919
goyalnarinder2 is correct.\n\nIf you also want to understand it, not only have the answer, then you should first know how to get rid of those brackets, for instance:\n\n(3x-1)(2x+3)= 3x(2x+3)-1(2x+3) \n\nGot the picture? It's called distributivity, the fact that you can distribute the factor 2x+3 to each of the terms of the first bracket - these would be 3x and -1.\n\nThen u continue distributing:\n3x(2x+3)=3x.2x+3x.3=6x^2+9x \n-1(2x+3)=-2x-3\n\nThat is how u get to:  \n6x^2+9x-2x-3=6x^2+7x-3 and this is just the result of the product of the first 2 brackets.\n\nAre u still w me? :-)\n\nYou now try to do the second product - which is (x-4)(x-1) - by yourself and see if you get this:\n(x-4)(x+1)=x^2-3x-4\n\nThen u add the 2 things (or rather substract...):\n(3x-1)(2x+3)-(x-4)(x+1)=(6x^2+7x-3)-(x^2-3x-4)\nand here u take care of the signs in the second bracket, which u should reverse (because say if I take -3 apples from u what I do is really give u +3 apples, got it?)\nSo u continue w:\nthe above = 6x^2+7x-3-x^2+3x+4=5x^2+10x+1\n\nHope this helped. Good luck!
823706
2x^2 + 13x + 39 remainder 122\n\ndoes no one know what synthetic division is?  i suppose i only learned it last year...
660785
i don't fucking know.\n\nEDIT: waitaminnit i'm gonna give it a try!\n\nx^3-4x^2-7x+10 = 0\n\nfor x = -1\n-1 -1 +7+10 != 0\nso x != -1\n\nfor x = 1\n1-1-7+10 != 0\nso x != 1 either!! ...goshdarnit.\n\nx^3-4x^2-7x+10 = 0\nx(x^2-4x) - 7x = -10\nx^2-4x - 7 = -10/x (we divided both sides by x)\n..ok...i think i've lost the plot sorry.
1079776
you can simply the first one to (x-a)(x-a) then you can divide that by (x-a) and it equals (x-a)\n\nDo you know synthetic division; cuz you can divide the second problem by (x+2)^2 if you mulitply (x+2)^2 out.
805033
(3x^2 + 5x + 1) / (4x-1) =\n\n3x^2 + 5x + 1/ 1  multiply to  1/ 4x - 1\n\nthe answer might be the same as the problem unless you want to derive it.\n\nthanks though
171139
whenever you divide numbers that have exponents, this rule will apply:\n\n(x^a)/(x^b) = x^(a-b)\n\nso in another words, applying your question:\n\n(x^3)/(x^7) = x^(3-7) = x^-4\n\nor 1/(x^4)
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.
821200
(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~ ♥ ~
741076
1/x+3/x+2=7\nMultiply by x getting:\n1+3+2x =7x\n4-5x =0\n0x* -5x +4 =0\n\n7/2x+2/x-1=4\nMultiply by 2x getting:\n7 +4 -2x =8x\n11-10x = 0\n0x* -10x +11 = 0\n\nx*+7x+2=2x+4\nx* +7x -2x +2 -4=0\nx* +5x -2 = 0\n\n4x*-3x=2x*+7 \n4x* -2x*-3x -7 =0\n2x*-3x-7 =0\n\n2x+3/x-1=x/4\nMultiply by x getting:\n2x* +3 -2x = x*/4\n2x* - x*/4 -2x +2=0\n2.25x* -2x +2 = 0\n\nThere, its done simply by multiplying and transposing terms to get what you want. You can move any term from one side of the equation to the other as long as you change its sign fro minus to plus or plus to minus.
1103929
(x^3+x^2-11x-15)/(x+3)\n= (x^3+3x^2-2x^2-6x-5x-15)/(x+3)\n={ x^2(x+3)-2x(x+3)-5(x+3)}/(x+3)\n={(x+3)(x^2-2x-5)}/(x+3)\n=x^2-2x-5\n\nThis should do. but on further simplification you will get \n   x^2 -2x -5\n={x^2 -1 -6^(1/2)}{x^2 -1 +6^(1/2)}
438371
same thing as (8x^2+2x-2)^(3/2), right?\nso add one to the power\nmultiply by the inverse of that power\nand divide by the derivative of what's in the parenthesis on the left\ngot it?
965349
I believe you meant (2x^2 - 3x - 2 )/ (10 + x - 3x^2) right?\n\nWell we can rearrange the problem to look like (2x^2 - 3x - 2 )/ (- 3x^2+ x +10)\n\nWe now factor out the top and factor out the bottom.\n[(2x +1) (x-2)] / [(-3x-5)(x-2)]\n\nSince this is a fraction, we can cancel out the (x-2) because it is on top and the bottom.\n\nWe are left with:\n(2x +1) / (-3x-5)\n\nI think this is what you wanted, good luck and be careful of how you write your notation!
902478
wow its been a long time.. i use a calculator to do that now.. but here goes one thing i remember...\n\nlets say we have a polynomial like this\n\nx^2 -3x-10  and we need to factor it..\n\nwe take a look at what factors of -10 we could use that would add up to -3  in this case there are two way to get -10 (-2*5) or (2*-5)  if we add the two sets together one gives us -3 and the other gives us 3 so we can throw out the -2*5 terms.\n\nSo now we lay out the factors.... we need two x's to get x^2 so we need two x's in the first terms (remember FOIL?)\n\nlay the terms out so they give us the correct values\n\n(x+2)(x-5)  now we can use FOIL (firs, outer, inner, last) to check ourselves...\nX*X = x^2\nX*-5 = -5X\n2*x =2x\n2*-5=-10\n\nso we get [x^2] +[-5x] +[2x] +[-10]\nor x^2 -5x+2x-10\ncombine like terms now\n-5x+2x =-3x\n\nwhich gives us\nx^2-3x-10\n\nhope that helps
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
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!