can someone explain me in brief what is "zero polynomial"?
465550
Do you mean "zero of a polynomial?" If so, then read on...\n\nA zero of a polynomial is the value or values of the variable (usually x) that cause the polynomial to equal zero.\n\nExample 1: The zero of the monomial 3*x is 0 because 3*0 is zero.\n\nExample 2: The binomial x + 2 has it's zero at x = -2 because if you substitute -2 in place of x, then you get \n---> -2 + 2 = 0.\nTherefore, -2 is the zero of the given binomial.\n\nExample 3: Find the zeros of the trinomial x^2 - x - 6. That is, solve x^2 - x - 6 = 0. If you factor the trinomial, then you get \n---> (x + 2)(x - 3) = 0. \nIn this problem there are two possibilities. Either \n---> x + 2 = 0  or  x - 3 = 0. \nSolving each gives x = -2 or x = 3. Therefore, the zeros of the given trinomial are -2 and 3. \n\nIf you substitute -2 into the given trinomial, then you have the following:\nx^2 - x - 6 \n= (-2)^2 - (-2) - 6 \n= 4 + 2 - 6\n= 6 - 6\n= 0 \n\nIf you substitute 3 into the given trinomial, then you have the following:\nx^2 - x - 6 \n= (3)^2 - (3) - 6 \n= 9 - 3 - 6\n= 6 - 6\n= 0 \n\nThere are no other values of x that will cause our given trinomial to equal zero.\n\nDefinitions\nA monomial is a polynomial with one term.\nA binomial is a polynomial with two terms.\nA trinomial is a polynomial with three terms.\nA polynomial with more than three terms is just called a polynomial.\n\nNote: If you graph a polynomial, then the graph will touch or cross the x-axis at the zeros of the polynomial. Zeros of a polynomial are also called solutions, roots, and, in some cases, x-intercepts.
153279
Zero (0) is a paradoxical number that has high philosophical and scientific importance. Context of zero in absolute & applied Mathematics is different. In applied mathematics we use it according to our convenience, for example 0 degree celsius, 0!(factorial), initial Velocity, initial position etc. And in Absolute Mathematics we think of it transcendently. Absolute zero is beyond our experience.Therefore we are using and thinking of zero in diffrent context. \nIn the fraction a/o, at present the numerator `a' is considered as something that is considerable. And the denominator `0' is not used in the absolute sense insted it  is taken in the concept of `infinitesimally small'.That is the symbol `0' here stands for `infinatesimally small'.For instance a `point' is considered as magnitude and dimensionless which is relatively true and wrong in the absolute sense.\nContext a/0=a is also correct iff same meanig is strected all through. ie. a/0 implies `a' is `undivided'.Then a*0 means a is not at all multiplied, hence a*0=a only in the absolute sense. Here `a' which is the quotient and  the product is same to both expressions implies that `a' is individual and Independent.This concept is non-pragmatic hence not in use.\nMathematically,a=(1+1+1...`a' times).\nGeneraly, if a/m =n, a=m*n = (m+m+m+...`n' times) \nAnd, if m is infinatesimally small,a=(m+m+m+... infinite  times).At present the parodoxical symbol `0' is used to denote the axiom `infintesimaly small'.consequently. it is written as a/0= `Infinity'\nNote that divisibility exists iff something exists in the denominator.And in the same way multiplicability exits iff something exits as multiplier.
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.
519094
Yes, zero is a number with a special property x + zero = x.\n\nIn most numeral systems, there is no zero symbol.\n\n0 is a digit who's name is zero.\n\n0 is not 'nothing'.  It is a number that can represent the cardanality of an empty set.\n\nThe greek capitol letter Phi (a circle with an up down line through it) is used to represent the 'empty set' or the set with nothing in it.
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.
448269
The domain is all real numbers x, except for x=0.\n\nYou just have to remember that the exponential function and all polynomial functions are defined on the entire real line, that is there domain is all real numbers. Therefore, the only problem that occurs is with the operation of division by x. Since you cannot divide by 0, x can never be 0. All other values are fine though.\n\nIN RESPONSE to what urtimud said below: sure you can take e^0, but if you let x=0 you are taking e^(-1/0) which is undefined. Therefore, x=0 is not in the domain of this real valued function. Note that e is being raised to the ((x^2-1)/x) power, because your screen may have chopped off that last part.
52943
it is a monomial\n\n\nMonomial is an algebraic expression consisting of a single term.\n\nBinomial is a polynomial with two terms. Example:(2x+3)\n\nTrinomial is a polynomial consisting of 3 terms.
579225
Division by zero is undefined, even if you divide zero into zero.\n\nhttp://en.wikipedia.org/wiki/Division_by_zero
146612
A quadratic equation is a second-order polynomial equation in a single variable  \n\n (1) \n\nwith . Because it is a second-order polynomial equation, the fundamental theorem of algebra guarantees that it has two solutions. These solutions may be both real, or both complex. \n\nThe roots  can be found by completing the square, \n\n (2) \n\n (3) \n\n (4) \n\nSolving for  then gives \n\n (5) \n\nThis equation is known as the quadratic formula. \n\nThe first known solution of a quadratic equation is the one given in the Berlin papyrus from the Middle Kingdom (ca. 2160-1700 BC) in Egypt. This problem reduces to solving \n\n   (6) \n   (7) \n\n(Smith 1953, p. 443). The Greeks were able to solve the quadratic equation by geometric methods, and Euclid's  (ca. 325-270 BC) Data contains three problems involving quadratics. In his work Arithmetica, the Greek mathematician Diophantus  (ca. 210-290) solved the quadratic equation, but giving only one root, even when both roots were positive (Smith 1951, p. 134). \n\nA number of Indian mathematicians gave rules equivalent to the quadratic formula. It is possible that certain altar constructions dating from ca. 500 BC represent solutions of the equation, but even should this be the case, there is no record of the method of solution (Smith 1953, p. 444). The Hindu mathematician Aryabhata (475 or 476-550) gave a rule for the sum of a geometric series that shows knowledge of the quadratic equations with both solutions (Smith 1951, p. 159; Smith 1953, p. 444), while Brahmagupta  (ca. 628) appears to have considered only one of them (Smith 1951, p. 159; Smith 1953, pp. 444-445). Similarly, Mahavira (ca. 850) had substantially the modern rule for the positive root of a quadratic. Sridhara (ca. 1025) gave the positive root of the quadratic formula, as stated by Bhaskara (ca. 1150; Smith 1953, pp. 445-446). The Persian mathematicians al-Khwarizmi  (ca. 825) and Omar Khayyám  (ca. 1100) also gave rules for finding the positive root. \n\nViète  was among the first to replace geometric methods of solution with analytic ones, although he apparently did not grasp the idea of a general quadratic equation (Smith 1953, pp. 449-450). \n\nAn alternate form of the quadratic equation is given by dividing (◇) through by : \n\n (8) \n\n (9) \n\n (10) \n\nTherefore, \n\n (11) \n\n (12) \n\n (13) \n\nThis form is helpful if , where  denotes much greater, in which case the usual form of the quadratic formula can give inaccurate numerical results for one of the roots. This can be avoided by defining \n\n (14) \n\nso that  and the term under the square root sign always have the same sign. Now, if , then \n\n (15) \n\n   (16) \n   (17) \n\nso \n\n   (18) \n   (19) \n\nSimilarly, if , then \n\n (20) \n\n   (21) \n   (22) \n\nso \n\n   (23) \n   (24) \n\nTherefore, the roots are always given by  and . \n\nNow consider the equation expressed in the form \n\n (25) \n\nwith solutions  and . These solutions satisfy Vieta's formulas \n\n   (26) \n   (27) \n\nThe properties of the symmetric polynomials appearing in Vieta's formulas then give \n\n   (28) \n   (29) \n   (30) \n\n \nGiven a quadratic integer polynomial , consider the number such polynomials that are factorable over the integers for  and  taken from some set of integers . For example, for , there are four such polynomials, \n\n   (31) \n   (32) \n   (33) \n   (34) \n\nThe following table summarizes the counts of such factorable polynomials for simple  and small . Plots of the fractions of factorable polynomials for  (red),  (blue), and  (green) are also illustrated above. Amazingly, the sequence for  has the recurrence equation \n\n (35) \n\nwhere  is the number of divisors of  and  is the characteristic function of the square numbers. \n\n Sloane factorable over  for , 1, ... \n A067274 1, 4, 10, 16, 25, 31, 41, 47, 57, ... \n A091626 1, 2, 4, 6, 9, 11, 14, 16, 19, 22, ... \n A091627 0, 0, 1, 2, 4, 5, 7, 8, 10, 12, 14, ...
589676
Nope.  Example: "1" is a polynomial "x" is also a polynomial (these are simple polynomials).  But "1/x" is not a polynomial.\n\nPolynomials divided by polynomials are always rational functions.
540459
something close to zero
444064
So, for a) & b) the two things that you should notice are:\n1 - The exponents inside the second polynomial alternate, meaning that as the powers of the x variable decrease, the powers of the y variable increase, both by increments of one.\n2 - The answer will only contain the variables x and y raised to the same power, which is one more than the leading exponent on the x variable.  \n\nSince (x + y)(x^2 - xy + y^2) = x^3 + y^3\n and  (x + y)(x^4 - x^3y + x^2y^2 - xy^3 + y^4) = x^5 + y^5,\nthen for any given polynomial the pattern is written out...\n(x + y)(y^2n - x^(2n-1)y + x^(2n-2)y^2 + ... + x^2y^(2n-2) - xy^(2n-1) + y^2n = x^(2n+1) + y^(2n+1)
513189
Neville's algorithm is an algorithm used to construct unique, interpolating polynomials. It uses n + 1 points to construct polynomials of degree n.\n\nIt is similar to Aitken's algorithm, which is nowadays not used.
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.
362894
Division by zero is undefined.
194771
1. Write down a couple of Jordan blocks.\n\n2. Write down their squares, cubes, maybe even a fourth power.\n\n3. Gaze at each set of identity matrix, J, J^2, J^3, ... until you see a simple linear combination of the matrices that sums to the all zero matrix.  That is your minimal polynomial.\n\n4. Generalize.\n\n5. Now it is obvious. :-)\n\nAlternative:\n\n1.  Google for some combination of:\n\nminimal polynomial Jordan block\n"minimal polynomial" "Jordan block"\nJordan block minimal polynomial\n"Jordan block" "minimal polynomial"\n\n2. Read answer.\n\n3. Too bad, it still isn't obvious. :-(
500175
Quadratic polynomials are usually in the form ax^2 + bx + c. They are usually factorable (which means you can find it's simplified roots). These roots should look like (x-F) and (x-G) where F and G are numbers so that when multiplied together using FOIL (First Outside Inside Last), you get your original polynomial. The roots are also where the polynomial = 0, assuming you've set it to zero. If the polynomials do not factor nicely, then you have to use this formula:\n\n-b +/- root(b^2-4ac)\n--------------------\n      2a\n\n\nWhich will give you the roots. Sometimes you get imaginary roots where in the equation above you will be taking the root of a negative number.\n\nHope that helps.