what are the simularities between radicals and rational exponents?
602511
radicals are root of a number and rational exponents are powers of a number.
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
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)
805031
1)  √10 · ³√100\n= 10^(1/2) · 10^(2/3)\n= 10^(7/6)\n\n2)  (6x)^(1/2) · (36x²)^(1/4)\n= (6x)^(1/2) · (6x)²^(1/4)\n= (6x)^(1/2) · (6x)^(2/4)\n= 6x\n\n...and so on.  Change the radicands to rational exponents with common bases, and follow the rules from there.  Good luck!
318167
It is possible to solve radical equations without graphing.
553656
No, a monomial is a type of polynomial.\nA monomial is a polynomial with only one term.\nPolynomials can only have non-negative integers as exponents.\n1/a = a^(-1)\nThe exponent is negative.\nThus, this is not a polynomial, which implies that it is not a monomial.
1042552
Here is my proof. Suppose P(x) is a polynomial in odd degree, for example 4x^5 + 3x^4 - 5x^3 + 2x^2 - 6x + 9. For large enough values of x, the term in x^5 will be so much larger than any of the other terms that it will swamp the smaller powers of x, and x^5 alone will determine the value of P. \n\nNow, if P is of odd degree and x is very large and positive, P will be positive; if x is very large and negative, then P will be negative. Therefore, somewhere between these two values of P there must be at least one point where P = 0. QED.
836453
a^24b^24\njust multiply the exponents\n\nhope thie helps!!!
446810
Let r be an element if R, and hence in L.  Since L is an algebraic extension of K, all elements of L, there is an irreducible polynomial:\n\n    P(x) = a0 + a1x + a2x^2 + ... anx^n\n\nwith the ai's in K such P(r) = 0.\n\nSubtracting a0 from both sides, we see:\n\n    -a0 = r ( a1 + a2r + a3r^2 .. + anr^(n-1))\n\nDividing both sides by -a0 we get (a0 is not zero because P is irreducible):\n\n      1 = r (-a1/a0 - a2/a0 r - a3/a0 r^2 ...)\n\nThe right side is in R since the the coefficients -ai/a0 are in K and r is in r.  So we have an inverse of r in R.
793079
The term with the highest exponent is x^5.  The coefficient of x^5 is one.\nThe constant term is 18.\nThe possible rational zeros are the factors of the constant term divided by each of the factors of the coefficient of the highest degree term.\nSince the coefficient of the term with the highest exponent is 1, that means the possible rational zeros are just the factors of the constant term.\nThe factors of 18 are 1, 2, 3, 6, 9 and 18.\nThe possible rational zeros could be negative or positive.\nAnswer: -1, +1, -2, +2, -3, +3, -6, +6, -9, +9, -18, +18
102907
Let:\nx = any number not equal to 0.\nm = any number\nn = any number\n\nProperty of exponents:\n\nx^m/x^n = x^(m-n)\n\nTherefore:\n1/1 =2/2 =3/3 = x/x = x^n/x^n = x^(n-n) = x^0 = 1
891987
The first step is to apply the Rational Roots Test to the polynomial to get a list of values that might possibly be solutions to the polynomial equation. You can follow this up with Descartes' Rule of Signs, if you like, to narrow down which possible zeroes might be best to check. Of course, if you've got a graphing calculator, it's a good idea to do a quick graph, since x-intercepts of the graph are the same as zeroes of the equation. Seeing where the graph looks like it crosses the axis can quickly narrow down your list of possible zeroes.\n\nOnce you've found a value you want to test, you use synthetic division to see if you can get a zero remainder. If you get a zero remainder, you've not only found a zero, but you've also reduced your polynomial by one degree. Remember that synthetic division is, among other things, division, so checking if x = a is a solution is the same as dividing out the linear factor x – a. This means that you should not return to the original polynomial for your next computation (for finding the other zeroes); you should instead work with the output of the synthetic division. It's smaller, so it's easier to work with.\n\nYou should expect to get some complicated solutions (that is, solutions containing square roots or complex numbers, or both). You will find these zeroes by applying the Quadratic Formula for the final quadratic factor.\n\nFind all the zeroes of the following polynomial: 2x5 + 3x4 – 30x3 – 57x2 – 2x + 24 \nFirst, I'll apply the Rational Roots Test--\n\nWait. Actually, the first thing I'll do is check to see if x = 1 or x = –1 is a root, because these are the simplest roots to test for. This isn't an "official" first step, but it can often be a timesaver, because you can just look at the powers and the numbers. When x = 1, the polynomial evaluates as 2 + 3 – 30 – 57 – 2 + 24 = –60, so x = 1 isn't a root. But when x = –1, I get –2 + 3 + 30 – 57 + 2 + 24 = 0, so x = –1  is a root, and I can do the synthetic division with x = –1 right now.\n\n\n\nThis leaves me with the smaller polynomial 2x4 + x3 – 31x2 – 26x + 24. (Since I've divided out the factor x + 1, I've reduced the degree of the polynomial by 1. That's how I know this is a degree-four polynomial.) Now I'll apply the Rational Roots Test to get a list of values to look at:
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
284432
There is no sign for exponent in your question.  I assume you mean\n2^(3/5)\nAny expression of the from x^(m/n) can be written as the\nnth root of x^m\nThus, 2^(3/5) is the fifth root of 2^3 which simplifies to \nthe fifth root of 8.
547302
Let radical 2 be a rational no. in the form of p/q.(p & q are int.,having no common factor & q iz not equal to 0)\nRdical 2 =p/q\n2 = p^2/q^2\np^2 = 2*q^2     (p^2 iz even, then p iz alzo even)\nLet p=2m\nThen p^2=4m^2\n     2q^2=4m^2      (p^2=2*q^2)\n     q^2 =2m^2   (q^2 iz even,then q iz even)\nBut p iz even & q iz even mean that 2 iz a common factor of p & q.Thi contradictz the above ztatement that p and q have no common factor.\nHence Radical 2  iz irrational.\n(PLZ- Ignore my "z'" az there iz an error in my board)
865479
First of all is the radical in the numerator over just the 10 or over the entire 10y^3?\n\nThe only this you can factor out and cancel is a radical 2 on the top and bottom.  Other than that, you can't really simplify this problem.  Are you sure you wrote it down correctly?
663667
It is true when both exponent & base are the same.\n\nOtherwise, they have to satisfy the formula:\n\na = r^(1/(r-1))\nb = r^(r/(r-1))\n\nwhere you can plug in any value for 'r' & get a, b - these are the two numbers we want.\n\n\nExamples:\n\n1. Suppose r=2,\n\na = 2^(1/(2-1)) = 2^1 = 2\nb = 2^(2/(2-1)) = 2^2 = 4\n\nAs you have shown yourself, 2^4 = 4^2.\n\n\n2. Suppose r=1.5,\n\na = 1.5^(1/(1.5-1)) = 1.5^2 = 2.25\nb = 1.5^(1.5/(1.5-1)) = 1.5^3 = 3.375\n\nYou can show using a calculator that 2.25^3.375 = 3.375^2.25\n\n\nP.S. Another very interesting property you will notice is that:\nb=(a*r) & b=(a^r)\n\nsimultaneously!