can someone explain me in brief what is "zero polynomial". please explain me the basic first. any help would be appreciated
210670
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.
98021
zero  you said squared, not to the zero power.  0^2=0  try it on your calculator.
264544
You simply divide each term in the polynomial by that integer.. For your example here it goes..\n\n(3x^2+8x-2) divided by 2 \n\n> {3(x^2)/2} + {8x/2} - {2/2}\n\n>> Answer: 1.5 x^2 + 4x -1    or   (3/2)x^2 + 4x -1
80892
There are actually two roots: When you factor it you get (x+2)(x+2)=0. So you have two roots: x=-2 and x=-2. It's just that, in this case, the two roots are called REPEATED roots because the same root occurs more than once.\n\nThis is in accordance with the fundamental theorem of algebra, since the polynomial is of degree 2.
410268
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.
97565
thanks 4 voting my answer as best but i didnt understand what u said\no divided by any no is always zero and with exception of zero where i guess it's not defined(aint that also zero iam not sure)
10436
Zero divided by zero is an indeterminate value. If you ever take a calculus course you will see this kind of thing crop up.\n\nThere are things you can do to evaluate expressions that are essentially exactly what you are asking about.....namely: zero divided by zero.\n\nThink about it.....you're saying you're going to arbitrarily call the result of a basic arithmetic calculation either zero or infinity. Two vastly different values. That alone signifies that more is needed in order to properly evaluate such a thing.
386466
The easiest function of x are:\n\nf(x) = constant\nf(x) = a*x, where a is a constant : 1st order poly\nf(x) = b*x^2 where b is a constant 2nd order\nf(x) = c*x^3 where c is a constant: 3rd order\n...\nf(x) = z*x^n, where z is a constant, n is a positive integer, nth order polynomial\n\na general polynomial is a sum of these.  They are convenient to work with for lots of reasons.  That's why they are so important.
536829
Im not going to prove it completely but I'll send you on your way..\n\nLook at a cubic, ax^3 + bx^2 + cx + d \nwe want to know for what value of x is this cubic = 0.\nLet's say r is a root.\nSet up the synthetic division:\n\nr || ..... a ........... b .............. c .................. d\n........................ ar .......... ar^2 + br ......... ar^3 + br^2 + cr\n───────────── ────────── ──────────\n......,....a......... ar+b ..... ar^2 + br + c......|| ar^3 + br^2 + cr + d\n\nin synthetic division, we say that if this last term,\nar^3 + br^2 + cr + d\nequals zero, then r is a root.\nNotice that the term is just the polynomial evaluated at x=r\n\nAnother way to look at it is:\nf(x) = ax^3 + bx^2 + cx + d\n= x( ax^2 + bx + c ) + d\n= x( x (ax + b) + c ) + d\n= x( x ( x (a) + b ) + c ) + d\nor,\ntake a,\nmultiply that by x, add b\nmultiply that by x, add c\nmultiply that by x, add d\nif that total is 0, then x is a root.\nThis is just what you do in synthetic division.
447488
Now my Algebra is a bit rusty but I think there is only one answe and it is {all reals}.   Here is my work...\n\n2x=x+x (subtract x from the right side)\n\n2x-x=x  (subtract x again)\n\n2x-2x=0\n\n0=0\n\nLike I said my Algebra is rusty, but I think I did that correctly.  The major thing I remember is to set one side equal to 0 then pick a variable to isolate.  We can do this another way too, I just noticed.\n\nx+x=2x therefore\n\n2x=2x \n\nThe answer is a set that contains all real numbers.
545203
An expression is undefined if denominator is zero.\n\nTherefore, 1.\nundefined when m-6=0  or m=6.\n\n2)  not clear\n\n3)y=0 not defined\n\n5)cancelling  (x-3)/(x+5)\n\nundefined when x=-5\n\n\n10)  -1\n\nnever undefined
273732
Division by zero is undefined, even if you divide zero into zero.\n\nhttp://en.wikipedia.org/wiki/Division_by_zero
595525
If one zero is 2-3i, another must be 2+3i, because complex zeroes always come in conjugate pairs. This means that a factor of P(x) is (x-2-3i)(x-2+3i)=x^2 - 4x +13.\n\nDivide P(x) by this and you get the third factor is (x-5), so the third zero is 5.\n\nTo find all the zeroes of f(x), using the rational roots theorem, the possible rational zeroes are +/-1, +/-2, +/-5, +/-10, +/-1/2 or +/-5/2.\n\nUsing synthetic division, fortunately it turns out that +/- 1 are rational zeroes. After factoring out (x-1) and (x+1), the reduced polynomial is 10x^2+x-2. That factors to (5x-2)(2x+1), so the remaining zeroes are 2/5 and -1/2.\n\nHope that helps.
78169
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. :-(
55151
(X-c) here represents the number you are going to devide with. c can take any value here.\nLet you have to devide with (x-5) , then c will be 5.\nAnd if you have to devide with (x+10) then c will be -10. As (x-(-10))=(x+10).\nThe remaider therum says, If you devide a polynomial, P(x) by (x-c) then the remainder will be P(c).\nIf you want a proof of this theorum,\nlet, the divisor of divison of P(x) by (x-c) is A and remainder is B.\nSo P(x)=A*(x-c)+B.\nSo P(c)=A*(c-c)+b\nor, B=P(c) which is the remainder, I hope this will clear your concept.
68726
two ways.\n1) this equation is equal to the product:\n(2x-1)*(-2x+1)=0\nSo, x=1/2\n2) For a polynomial of degree 2, in the form of \n((a*x^2)+(b*x)+c).the discriminant is [(b^2)-(4*a*c)].Here, a=-4 b=4 c=-1.\ndiscriminant= 16-(4*(-4)*(-1))=0\nthe formulas for x are =[-b+((dis)^(1/2))]/(2*a) and\n[-b+((dis)^(1/2))]/(2*a).\n\nSo, x=[-(-4)+0]/(2*4)=1/2 and =[-(-4)-0]/(2*4)=1/2
187748
0! = 1 is a definition according to my math textbooks.\n\nYou can also look at it this way.  There is only 1 way to choose 0 objects.