rational expressions and their applications are like fractions right..?
551937
any number in the form p/q where p,q belong to Z(integers) and q is non-zero, p/q is called a rational number.\nall fractions are rational numbers.\nthese can be expressed in decimal form as well.\nAs you say, operations on rational expressions and their applications are like those of fractions.\nAddition and subtraction, multiplication, division of two or more expressions is like those of fractions.\n\n2/3+5/6=?\nfind the LCM of 3 and 6. it is 6.\n4/6+5/6\n=(4+5)/6\n=9/6\n\nsimilarly,\na/b+c/d can be solved.
291495
Rational, because it can be expressed as the quotient of 2 integers:\n-2 5/8 = -21/8\n\n^_^
721902
can you elaborate on that? what do you mean by simplify decimals? you can put decimals into fraction form and simplify that.
729449
It will not factorise in the rational number field so there are no rational solutions\n\nIrrational roots are -1.72107 and 0.269209 (approximately) using Quickmath
970218
Do you require mathematical signs, or are standard operations OK (like square root)?\n\n1: (1 + 1 + 1)!\n2: 2 + 2 + 2\n3: 3 * 3 - 3\n4: sqrt(4) + sqrt(4) + sqrt(4)\n5: 5 + 5/5\n6: 6 + 6 - 6\n7: 7 - 7/7\n8: 8 - sqrt(sqrt(8 + 8))\n9: 9/sqrt(9) + sqrt(9)\n\nAnd for a bonus: (0! + 0! + 0!)! = 6
557964
Wouldn't it be something like:\n\n2x(8) + 8s     to get rid of the denominator\n\n16x + 8s\n\nThat is as far as I think you could simplify it, because each has different variables and you can't combine something with different variables.
685531
[(2x^3+2x^2+3x+3)/(x^2-9)]*\n[(x^2+2x-15)/(4x^2+6)]\n\nFactor each expression individually.\n\n(2x^3+2x^2+3x+3) = (x+1)(2x^2+3)\n\n(x^2-9) = (x+3)(x-3)\n\n(x^2+2x-15) = (x+5)(x-3)\n\n(4x^2+6) = 2(2x^2+3)\n\nSubstituting these in, our expression becomes\n{[(x+1)(2x^2+3)]/[(x+3)(x-3)]}\n*{[(x+5)(x-3)]/[2(2x^2+3)]}\n\nCombining these into one larger expression, we get\n[(x+1)(2x^2+3)(x+5)(x-3)]/\n[2(x+3)(x-3)(2x^2+3)]\n\nCancel the quantities common to the numerator and denominator and we get our final answer\n[(x+1)(x+5)]/[2(x+3)]\n\nGood luck.
733179
There are an infinite quantity of rational numbers between -1 and 1.\n\nFor example all of these are rational numbers in this range:\n-1, -7/8 -3/4, -5/8, -1/2, -3/8, -1/4, -1/8, 0, 1/8, 1/4, 3/8, 1/2, 5/8, 3/4, 7/8, 1\n\nBasically anything in the form a/b where a and b are integers, and the ratio of a/b is between -1 to 1 will satisfy the equation.\n\nGiven integers a and b, where 1 < a < b -- any of these will be valid rational numbers in the range -1 to 1:\n\na/b, 0, -a/b
72448
The set of all rational numbers are countable in the following pattern..\nhttp://www.homeschoolmath.net/teaching/rational-numbers-countable.php\nNow from this you can easily verify if you multiply all the numbers by 2/3 then you will get the set you want..So this set is also countable..
303592
Yes, I can solve them.\n\nLike any other fractions, find a common denominator.  Though with (a) you can just multiply everything by 7x and simplify the equation to:\n\n21 + 1 = -14x\n\nYou can do the same with (c).  With (b) it's a little more complicated, and you have to find the common denominator of x and (x + 6).  I'll leave that for you.
415757
I hate to tell you, but I don't think there is an algebraic method for arriving at a value for x. Sometimes, this can be the case, in which case all you can do is find a numeric approximation, here around x=1.0587...\n\nI'll see if another user can prove me wrong, but I'm going to go with there's no way to solve this problem using analytic methods.
715426
x^4 + 5x^3 - 30 / x^2 - 3\n\nThe important thing is that, especially with division, polynomials need place value so 0 coefficients apply.\n\nRewrite as follows\n\nx^4 + 5x^3  + 0x^2 + 0x - 30 / x^2 + 0x - 3\n\n.................... x² + 5x + 3\n................. ____________________\nx² + 0x - 3 )x^4 + 5x³ + 0x² + 0x - 30\n.................. x^4 + 0x³ - 3x²\n.................  ------------------ .....↓\n........................... 5x³ + 3x² + 0x\n........................... 5x³ + 0x² - 15x\n........................... ------------------ ....↓\n.................................... 3x² + 15x - 30\n..................... ............... 3x² + 0x - 9\n.................................... ------------------\n....................... ................... 15x - 21\n\nSo x^4 + 5x³  - 30 = (x²  - 3)(x² + 5x + 3) + 15x - 21\n(Or quotient is x² + 5x + 3 and the remainder is 15x - 21)
1123203
You can multiply and divide them easier.\n\nfifth root of 2x times 2x^20\n           -or-\n2x^(1/5) times 2x^20
791694
The rational roots of any polynomial:\n\na(n)x^n + a(n-1)x^(n-1) + ... + a(1)x + a(0) \n\nare in the form ±p/q, where p divides a(0) and q divides a(n).\nIn your case, a(0) = -1 and  a(n) = 1, so the only rational roots possible are ±1.\nSubstitute ±1 into the equation to see if either is a root. (neither is)\n\nExample:\nx^3 + 4x^2 + 2x - 4 = 0\nThe only possible rational roots are ±1, ±2, ±4 (These numbers divide a(0), and only 1 divides a(3)=1 ).  Substituting each of these into the equation (synthetic substitution works well), we find -2 the only rational root.\n\n*****************************************\nProof by contradiction.\nLet the rational number p/q be a reduced root (i.e, q  and p are relatively prime integers) to the rational polynomial equation:\na(n)x^n + a(n-1)x^(n-1) + ... + a(1)x + a(0) = 0\nand such that either p does not divide a(0) or q does not divide a(n).  [Note: without loss of generality we can assume all coefficients are integers by multiplying both sides by the least common multiple of the denominators of all the a(n)'s]\n\nThen we know that \na(n)p^n/q^n + a(n-1)p^(n-1)/q^(n-1) + ... + a(1)p/q + a(0) = 0\nmultiply both sides by q^n to get:\n\na(n)p^n + a(n-1) qp(n-1) + a(n-2) q^2p^(n-2) + ... + a(1) q^(n-1)p \n+ a(0)q^n = 0\n\nor a(n)p^n + q * f(a,p,q) = 0 for some function f of a,p,q.\nand a(0)q^n + p * g(a,p,q) = 0 for some function g of a,p,q.\n\nThis means \na(n)p^n = -q * f(a,p,q)\na(0)q^n = -p * g(a,p,q)\n\nBy the first equation, q must divide a(n) * p^n.\nBut q has no factor in common with p, so it cannot have a factor in common with p^n, so all the factors of q (ie, q itself) must divide a(n), and we assumed that q does not divide a(n), so p must divide a(0) * q^n (the only possibility left)\nBut likewise with the second equation, p must divide a(0) * q^n. But p does not have any common factors with q, so it cannot have common factors with q^n, so p must divide a(0) and we assumed that p does not divide a(0), so a contradiction arises.\nTherefore no such p/q can exist, therefore the roots are irrational.\n\nQ:\nwhy shouldn't q divide a(n) or p divide a(0)?\n\nBecause otherwise if both did then the roots COULD be rational (p/q could be a root).  If thats the case you prove/disprove it by example.  In other words, its easy to show that in your example the roots are all irrational because the only POSSIBLE rational ones are 1 and -1.  So you just substitute them in to see that they're not roots. My proof just shows that IF there are rational roots, they must be p/q where p|a(0) and q|a(n).\n\nThere is no general way to determine  by inspection that a random polynomial has all irrational roots. The only way is to check the POSSIBLE rational ones by substitution.
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.
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
221137
Well, 2/5 is already a fraction.  It is a "rational fraction," which means that it is expressed as a ratio, i.e. one number divided by another.\n\nYou can also express it as a decimal fraction.  Divide 5 into 2 by the usual means you do division and you will find that 2/5 = 0.4.\n\nYou can also express it as a percentage.  Just multiply the decimal fraction by 100: 0.4 x 100 = 40%.