This is dealing with dividing monomials and simplifying expressions with negative exponents.?
445985
if by this you mean\n\n((3x^-2 * y)^-2)/((4xy^-2)^-1)\n\nthis becomes\n\n((3^-2) * x^(-2 * -2) * y^-2)/((4^-1) * x^-1 * y^(-2 * -1))\n\n((1/9)x^4 * y^-2)/((1/4) * x^-1 * y^2)\n\n((1/9)/(1/4)) * x^(4 - (-1)) * y^(-2 - 2)\n\n((1/9)*(4/1)) * x^(4 + 1) * y^(-2 + (-2))\n\n(4/9)x^5y^-4\n\nwithout using negative exponents, the answer would be\n\n(4x^5)/(9y^4)
1079811
It *is* awfully hard to write algebra expressions on Yahoo Answers, isn't it? Oh well, we do what we can.\n\nAddressing your first question, let's write it out like a long division problem.  You'll have to imagine the horizontal bar above the larger polynomial, because I can't format it properly here:\n\nx^2 + 3x - 2  )  x^4 + x^3 - 5x^2 + 13x - 6\n\nOur first step is to look at the first term in each polynomial, and divide the smaller into the larger.  x^4 / x^2 = x^2, so the first term we write in our quotient (above the bar) is x^2.  You can write it directly above the x^4, if you wish, or above the 5x^2 term, your preference.\n\nNow, we're going to multiply that "partial quotient" by the entire divisor (x^2 + 3x - 2), which gives us x^4 + 3x^3 - 2x^2.  Write that "partial product" below the first three terms of the original dividend (x^4 + x^3 - 5x^2).  Our next step is to subtract the partial product: (x^4 + x^3 - 5x^2) - (x^4 + 3x^3 - 2x^2) = -2x^3 - 3x^2.  We then drop down the next term from the dividend, the 13x term, and repeat:\n\n-2x^3 / x^2 = -2x (add that term to our quotient)\n\n-2x * (x^2 + 3x - 2) = -2x^3 - 6x^2 + 4x\n\n(-2x^3 - 3x^2 + 13x) - (-2x^3 - 6x^2 + 4x) = 3x^2 + 9x\n\nFinally, we drop down the last term from the dividend, -6, and repeat once more.  I'm sorry that I can't write this out the way your math book can... I really hope Yahoo Answers creates some sort of way of inputting formulas, or at least monospacing text entry!\n\n3x^2 / x^2 = 3 (so we add a +3 term to our quotient)\n\n3 * (x^2 + 3x - 2) = 3x^2 + 9x - 6\n\n(3x^2 + 9x - 6) - (3x^2 + 9x - 6) = 0... no remainder!\n\nSo our final quotient is x^2 - 2x + 3, with no remainder.\n===\nNow, when doing the 2nd problem, you'll want to write the larger polynomial, the dividend, as x^4 - 3x^3 + 0x^2 + 6x - 5.  We incorporate that extra 0x^2 term so that we never find ourselves trying to subtract unlike terms.\n\nSorry there's no better way to write this out on Yahoo Answers.  Hope that helped anyway!
1010165
symplifying an expression is breaking it down.  This is a very simple example: 3x=9 what you do is divide each side by 3 so you're left with  x=3.
452207
Math teacher here...\n\nParentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right).  That is the order of operations.\n\n1) -[-11+6]   Must carry out the parentheses first.  -11+6 = -5, so -(-5) is 5.\n\n2) -[10+(-4)]  Must carry out [10+(-4)] = 10 - 4 = 6.  So, -(6) = -6.\n\n3) 3[-2] + [-3]  Everything in parentheses is simplified, and there are no exponents, so on to multiplication.  3[-2] = 3 * (-2) = -6.  -6 + (-3) = -6 - 3 = -9.
1010071
ok its like this...() or parenthesis meant multiplication. Then multiplyin negative by negative is = to +. So now its gonna be +a or simply a
843761
well assuming you just want to break it down...\n1: you put the -2 power on both the 4 and a^3.  then it'd be (4^-2)(a^3^-2).\n2: remember that when it's an exponent to an exponent you multiply the numbers together so it becomes (4^-2)(a^-6)\n3: a negative exponent means it is one over itself so it would be 1/(4^2)(a^6)\n4: just simply from there to make it 1/ (16a^6)\n\nhopefully that was what you were asking!!
759262
The rules for simplifying integer exponents are pretty simple:\n\nhttp://en.wikipedia.org/wiki/Exponentiation
66583
Polynomials are expressions involving x raised to a whole number power (exponent). Some examples are: \n\n\n\nIn this lesson we consider division of polynomials such as: \n and . \nThere are two ways to calculate a division of polynomials. One is long division and a second method is called synthetic division. Synthetic division can be used when the polynomial divisor such as x-2 has the highest power of x as 1 and the coefficient of x is also 1. Otherwise, such as , a long division process must be used. Both methods will be demonstrated. \n\nWe will now look at some examples showing how these two procedures are done
824178
You typed exponents just fine!\n\nx^3 - x^2 -x + 7 divided by x+1 is a "long division" type problem rather than factoring because x+1 doesn't go evenly into the polynomial\n\nThe answer is x^2 -2x + 1 + 6/x+1
8373
(([(4+4+4).(4+4)]+4)/4)*(4+4)\n\n(4+(4*4)).([(4+4+4).(4+4)]+4)
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
154296
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.
196269
i took this question to mean as:\n\n[11/ (4-x)] + [2x/ (16-x^2)]\n\nso, what you do first is to multiply addends by (-1/-1), which is still equal to one, and won't make a difference, except to change the signs.\n\nso it would look like\n\n[-11/ (x-4)] + [-2x/ (x^2 - 16 )]\n\nsimplify to:\n\n[-11 / (x-4) ] + [ -2x / (x+4) (x-4) ]\n\ncombine the elements:\n\n[(x+4) + (-2x)]/ [(x+4) (x-4)]\n\nto make \n\n[(x+4-2x)]/ [(x+4) (x-4)]\n[(-x+4)]/ [(x+4) (x-4)]\n[-(x-4)]/ [(x+4) (x-4)]\n\nand the simplest would be:\n\n-1 / (x+4)\n\n hope this helps. :D
381493
It's set up like a long division problem.  Let's say you have x^2 - 6x - 27 divided by x + 3.  You'd set up like this:\n\nx + 3 ) x^2 - 6x - 27\n\nFor polynomial division, just focus on the variables and the exponents.  To start this one, look at how many times x goes into x^2: x times.  So x would go on top of the dividing symbol, and then you'd multiply the value on top by the value on the outside and subtract, just like a long division problem.  So you'd end up with\n\nx^2 - 6x - 27\n-(x^2 + 3x)\n----------\n-9x - 27\n\nRemember that, just like long division, you're subtracting, so the problems becomes x^2 - 6x - 27 - x^2 - 3x if you wrote it out this way.  The x^2 terms cancel out, and you'd have to bring down the -27.\n\nNow repeat the process - how many times does x go into -9x?  -9 times.  So the -9 goes on top (making your top x - 9) and your problem look like this:\n\nx^2 - 6x - 27\n-(x^2 + 3x)\n----------\n-9x - 27\n-(-9x - 27)\n----------\n0\n\nSince you have no remainder, your answer is x - 9.  And if you're ever unsure, you can always use FOIL to see if you get your original trinomial:\n\n(x + 3)(x - 9) = x^2 - 9x + 3x - 27\n= x^2 - 6x - 27
812516
Concentrate on the ^2, because that will have the most significant impact on the value of the expression being raised to the second power:\n\nFirst I tried just raising the 4 to the second power.  4^2 = 16, so "n" / 16 = 6, therefore n = 96.  If you can group the other numbers to reach 96, you have an answer.  But working that out for a while got me nowhere, so I started working leftwards to bring the next number and operator into the square:\n\n(6/4)^2 = 2.25\n\nThat is not an integer so it probably won't work.\n\nEven bringing the next number and operator in, which is 2 -, I still end up with a non-integer:\n2 - (6/4)^2 = -0.25\n\nBut look -- the next number and operator are "8 x " so lets work that out:\n\n8 x (2 - (6/4)^2) = 8 x (2 - 2.25) = 8 x (-0.25) = -2\n\nso that might work if we can come up with a number which works out for ___ + (- 2) = 6\n\nWhat plus -2 is =6?  It is 8 of course.  Now we just need to group 3 x 5 - 3 + 2 to equal 8:\n\n(3 x (5-3)) + 2\n(3 x (2)) + 2\n(3 x 2) + 2\n6 + 2\n8\n\nAha!  We have it.\n\nSo then, or grouping is:\n(3 x (5-3)) + 2 + (8 x (2 - ((6 / 4) ^2))) = 6\n\nTechnically some of those parenthesis are extras because the order of operations says multiplications and divisions are always done first, you don't need to explicitly show that they are (but it helps keep things clear).  So this would be a simpler form:\n3 x (5 - 3) + 2 + 8 x (2 - (6 / 4) ^2) = 6\n\nBut anyway, putting it back in the more complex form and proving it true works like this:\n(3 x (5-3)) + 2 + (8 x (2 - ((6 / 4) ^2))) = 6\n(3 x (5-3)) + 2 + (8 x (2 - ((6 / 4) ^2))) = 6\n(3 x ( 2  )) + 2 + (8 x (2 - (( 1.5 ) ^2))) = 6\n(3 x   2   ) + 2 + (8 x (2 - (    2.25   ))) = 6\n(3 x   2   ) + 2 + (8 x (      -0.25        )) = 6\n(6) + 2 + ( -2) = 6\n6 + 2 - 2 = 6\n6 = 6 <<<<< TRUE!\n\n\nHere is another grouping that also works out:\n3 x (5-3) + 2 + 8 x ((2 - 6) / 4^2) = 6\n3 x    2    + 2 + 8 x     (-4 / 16) = 6\n3 x    2    + 2 + 8 x (-0.25) = 6\n3 x    2    + 2 - 2 = 6\n6 + 2 - 2 = 6\n\nand another, this one involves only integers in the proof so might be more what you are looking for:\n3 x (5-3) + 2 + (8 x (2 - 6)) / 4^2 = 6\n3 x 2 + 2 + (8 x (-4)) / 16 = 6\n3 x 2 + 2 + (-32) / 16 = 6\n3 x 2 + 2 + (-2) = 6\n3 x 2 + 2 - 2 = 6\n6 + 2 - 2 = 6\n6 = 6
42795
178 is a monomial....\n\nmonomial are those which has only 1 expression... eg 12, x,125\n\nbinomial - 2 expressions eg 12x+45y\n\ntrinomial - 3 expressions eg 12x+45y+78z
888747
I think you meant 3^0 + 3^1 + 3^2 + ... + 3^(n-1)\nNote the parenthese.\nIt reduces to (3^n - 1) / 2\n\na^0 + a^1 + a^2 + ... + a^(n-1) reduces to (a^n - 1) / (a - 1)