I need to have a proof that shows why complex roots of a polynomial occur in conjugate pairs.?
923173
You have to assume that all the coefficients are real numbers. Then, if z satisfies\na_n z^n+...+a_1 z+a_0=0,\ntake the complex conjugate of the equation to get\na_n (z*)^n+..a_1 z* +a_0=0.\nIn other words, if z is a root, so is z*, so roots occur in conjugate pairs.
54176
The instruction is to write a polynomial function with the zeros 0, 1, and 1+i. Please keep in mind that when a zero consists radicals or imaginary numbers, the conjugate of that zero is also a zero. For exmaple, here 1+i is a zero. Then 1-i must also be a zero.\n\nThis problem is like working backwards. Instead of factoring the polynomial, you are asked to do it backwards, starting with the zeros. Since 0, 1, and 1+i and 1-i are zeros, you would expect the last step of the factoring to look something like this:\n\nx(x-1)[x-(1+i)][x-(1-i)]\n\nNow simplify this expression a bit. It becomes:\n\nx^2-x[(x-1-i)(x-1+i)]\n\nSimplify this expression even further and you get:\n\nx^4-3x^3+4x^2-2x
894899
Horners' method of synthetic division is used to find the roots of polynomials. \nwe have to know one of the roots to proceed with this. we can check if 1or -1 are roots of the polynomial using some easy tricks. \n1) if the sum of all the coefficients is equal to 0, then 1 is a root\n2) if the sum of coefficients of even terms is equal to the sum of the coefficients of the odd terms, then -1 is a root.\nnow, for the first equation, we already know that -2 is a root, because the question is to find the quotient when the polynomial is divided by x+2. \nin synthetic division, u first put down a 0 beneath the first coefficient, ie., 3. now add these two. the result is 3. \nnow multiply 3 with the root, ie., -2. the result is -6. write this beneath the second coefficient, ie., 2. the result is -4 after addition. \nrepeat the same steps until u reach the last coefficient.\nnow next we write 8 below -1. result is 7.\nnext we write 7*-2 ie., -14 below 5. the result is -9.\nlastly, we write -9*-2=18 below . the result is 12, but the result should be zero.\nso, the given equation is not divisible by x+2.\n\ncoming to the second equation, follow the same procedure as above and u will get the resultant coefficients at each stage to be 2,6,22,61 respectively. so it is not divisible by 3
653235
It's not square *roots*... it is the series of PERFECT SQUARES!!! How many times are people going to ask this? And isn't it obvious? This is part of some Google DaVinci Code contest, isn't it?\n\nWhy don't you just search for the same question because it has been asked about 31 times now!!!
498942
what proof are you looking for? can you specify more details about the actual question?
460547
This looks close enough to the techniques used to show that the roots are close to the critical line. The actual estimates are difficult, though, and nobody has been able to actually give a proof using these techniques (obviously). Go ahead and try it. Perhaps you can see through to a proof. If so, you will be famous. If not, you will have learned somthing.
742818
simple just find out DISCRIMINANT= b^2 - 4ac\n\nif Dis is +ve then there are two real distinct roots\n\nif Dis is 0 then there is exactly one root.\n\nIf the discriminant is negative, there are no real roots. Rather, there are two distinct (non-real) complex roots, which are complex conjugates of each other\n\nin your question\n\ncheck for solving quintic equations and  bring it in the form of \nquadratic equation.
401494
From,\n(3-5/2)^ 2 = (2-5/2)^ 2\n\nThe mistake is in \n"Taking positive square root on both sides"\n\nLHS = Left Hand Side\nRHS = Right Hand Side\n\nFor we cannot simply take square roots. \nLook at LHS (3-5/2) is the POSITIVE root. \nOn the RHS (2-5/2) is the NEGATIVE root.\nYou made them "equal" to get the "wrong proof".\n\nActually, to take the roots note that \n(2-5/2)^ 2 = (5/2 - 2)^ 2\nSo, If you wanted to take "positive root" then,\nits obviously (5/2 - 2).\n\nThus comparing positive roots,\n3 - 5/2 = 5/2 - 2 which is a identity.\n\nRegards
706946
In general it is difficult to solve a cubic equation; there is an algebraic formula for the solutions of a general cubic, but it is extremely complicated.\n\nTry to find a solution by inspection. First I will use the rational-roots test, which is that all rational roots of a polynomial (if they exist) can be written in the form a/b, where a divides the constant term and b divides the leading coefficient. This means that the only possible rational roots are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60, and their negatives. It is not hard to see that none of these is a root.\n\nNow I will try elements of the extension fields of the rationals by square roots of factors of 60.\n\nIf x=a+b√2, then a^3+3a^2*b√2+6ab^2+2b^3√2+2a+2b√2=60, so a^3+6ab^2+2a=60, and 3a^2*b+2b^3+2b=0, so 3a^2+2b^2=-2, which is impossible if a and b are rational; similar things happen with any other extension fields of degree 2.\n\nYou can also try extension fields of higher degree to see whether you get lucky.\n\nAnother idea is to find two values of x such that x^3+2x-60 is negative for one value and positive for another (for example, 3 and 4), and use procedures like Newton's method or regula falsi to approximate the real root (a graph indicates that there is only one real root).\nI cheated with my calculator and found no easy factorization.
750403
I have to go teach my class in about 4 minutes, but when I return, if no one has done it, I can show you how.  It requires factoring out a specific value that makes the coefficients "a" and "b" turn into two values that satisfy x^2 + y^2 = 1.  In other words, I know it starts out by factoring out sqrt(a^2 + b^2) from \n\na sin(Bx)+b cos(Bx)\n\n\nto get \n\n\nsqrt(a^2 + b^2) [ a/sqrt(a^2 + b^2) * sin (Bx) + b/sqrt(a^2 + b^2) * cos(Bx)]\n\n\nEDIT: Okay here we go.  And the "orthogonal rotating vectors" proof below is usually over most people's heads, or at least, most people who would ask this question on Yahoo answers.\n\n\nFirst, you factor out SQRT(a^2+b^2) as I mentioned above.  For the sake of simplicity, I will let A = SQRT(a^2+b^2).  So factor A out of  \n\na sin(Bx)+b cos(Bx)\n\nto get\n\nA [a/A sin(Bx) + b/A cos(Bx)]\n\n\n\n\nSince (a/A)^2 + (b/A)^2 = 1 (you might need to actually verify this for your purposes), there exists some angle C such that\n\ncos (C) = a/A and sin (C) = b/A \n\n(also, you should point out that both a/A and b/A have absolute values less than one, so that C is real and not complex with a nonzero imaginary part).\n\nSo make the substitutions to get\n\nA [cos(C) sin(Bx) +  sin(C) cos(Bx)].\n\n\n\n\n\nAccording to the formula \n\nsin(y+z) = sin y cos z + cos y sin z, \n\nthe expression in the brackets must equal sin(Bx + C).  So now the whole expression becomes\n\nA sin (Bx + C).\n\n\n\n\n\nSo in conclusion, \n\na sin(Bx)+b cos(Bx) = Asin(Bx+C),\n\nwhere A = SQRT(a^2+b^2), and C = arccos(a/A).  \n\n\n\n\nThere is a better representation for C, however.  \n\nSince cos C = a/A and sin C = b/A, you have\n\ntan C = (sin C)/(cos C) = (b/A)/(a/A) = b/a.\n\nThus, tan C = b/a, and C = arctan (b/a).  This definition of C does not rely on the creation of a third variable (A).\n\n\n\n\n\nHope this helps!
271648
boy we carry those text books along with us\nif u want us to prove then provide the theorem plz
201685
If y is a root of this and b is in F_p, then (y+b)^p -(y+b)=y^p +b^p -y-b=a, so y+b is another root. Hence the complete list of roots is y, y+1, ....y+p-1.\n\nThus we have two cases, the first where the polynomial splits in F_p (a=0) and the other where we get a degree p extension, so F_(p^2).\n\nNotice that we are using the algebra student's dream (x+y)^p=x^p +y^p which is justified because we are working over a field of characteristic p.
474735
If z = x+iy, then the complex conjugate of z is Z'= x-iy.\nIn your sum the conjugate of 2ix - i = i(2x-1).  The real component is zero and the imaginary component is 2x-1. Then the conjugate of 2ix - i must be i(1-2x).  Similarly, \nin (1-2x) the real component is 1-2x and the imaginary component is zero.  The complex conjugate is 1-2x.
467967
this is probably not a perfect answer but this site may give you some clues.\n\nhttp://www.gomath.com/Questions/question.php?question=4751\n\naltrnatively type 'prove v*v = 2gr*r/s+v0*v0-2gr' into a search engine and see what comes up.
973435
A polynomial is said to be irreducible if it cannot be factored into nontrivial polynomials over the same field. \n\nover real number x^2+1 is irreducible but not in complex number domain. This can be exteded to modulus. Please refer to
551184
its simple distributive law.\nwhat has happened is they have taken (a-2) common out of both the terms .\neg: cx-x = x(c-1)\nthe same thing has happened here with x=(a-2) and c=a^2\nso, in step 2, we have: a^2(a-2)-(a-2)\ntaking (a-2) common outside; you end up with (a-2)*((a^2)-1)\nwhihc is [3]\nthen as you know, (a^2)-1 is same as (a^2)-(1^2) and then we apply the algebraic identity (a^2)-(b^2)=(a+b)*(a-b) to get (a-2)(a+1)(a-1) which is [4]
411931
Neither answer is correct.  \n\nExtraneous roots appear when solving an equation in a perfectly correct way, becuase certain operations have more than one solution.  Taking square roots, for example, gives TWO answers, cube roots give three, etc.  (ESPECIALLY when we use complex numbers.)  This isn't a mistake, just part of the solution process.  \n\nIt is necessary to check results against the original problem whenever these operations are used.