Reducible   polynomial?
441742
Assume there is a pair a_i, b_j with a non zero product.\nWrite fg= (a_i x^i + other terms) (b_j x^j + other terms) =\na_i b_j x^(i+j) is non zero.\n\nContradiction.
461122
Since f(x) is a polynomial of degree two in x,\nLet f(x) = ax^2 + bx + c, where a,b,c are constants.\n\nLet's find f(x) first.\n\nf(0) = 1 implies, c = 1.\n\nf(1) = 3 implies, a + b + c = 3\nor  a + b = 2 , or b = 2 - a  ...(1)\n\nf(-1) = 5 implies, a - b + c = 5\nor a - b = 4 ...(2)\n\nSubstituiting (1) into (2), we get,\n\na - [ 2 - a  ] = 4\n2a = 6 or a = 3.\nThus, b = 2 - a = 2 - 3 = -1.\n\nHence, f(x) = 3x^2 - x + 1\n\nNote that by long division or by the identity below,\n\n3x^2 - x + 1 = 3(x + 1)^2 - 7x - 2\n             \nWe get,\n\n[ 3x^2 - x + 1 ] /[1 + x]^2 = 3 - [7x + 2]/ [1 + x]^2   \n\nAnd that together with below,\n\n[7x + 2]/ [1 + x]^2  = 7 /(x + 1) - 5 /(x+1)^2\n\nGives,\n\n[ 3x^2 - x + 1 ] /[1 + x]^2 = 3 - 7 /(x + 1) + 5 /(x+1)^2\n  \nNow, we have an "integratable form".\n\n∫ [f(x) / (1 + x)^2] dx\n\n= ∫ [3x^2 - x + 1] / (1 + x)^2] dx\n\n= ∫ [ 3 - 7 /(x + 1) + 5 /(x+1)^2 ] dx\n \n= 3x - 7 log(x + 1) - (5/[x + 1]) + C\n\nwhere C is any arbitary constant.\n\nNote: Sometimes log(x + 1) can also be written as ln (x + 1)\n\nRemark. It is generally accepted that log and ln are both in base "e", the Euler number. If its in a different base, then usually its said or stated. For example logarithm of x in base 2 can be written as log_2(x).\nJust symbols :)\n\nRegards
806827
depends what your trying to sum from the polynomial... what are you trying to get, theres to many thing just to say sum of given polynomials
922060
My suspicion upon seeing this question is that it can be solved using mathematical induction.  After all, couldn't we simply 'build' such a (maximally connected) graph starting with n=2 vertices and 'grow' it from there?\n\nRather than take on the proof myself -- mostly because I am unsure of the definition of 'components' -- I leave it as an exercise for you.  However, it bears noting that the formula shown is the commonly accepted "sum of the first n integers" formula used in almost every text on induction.\n\nGood luck!
890098
I enjoy algebra, and this is easily my favorite type of problem.  So I hope you, too, one day enjoy them for their inner beauty.\n\nFor now, let's tackle this "factoring problem."  We may wish to use the method of "synthetic division" which is arguably easier than "long division of polynomials."\n\nThink back to grade school where you learned that 6 = 2 * 3.  This simple statement enabled you to 'deconstruct' the number 6 as either "two times three" or "three times two".  Even if you forgot one of the 'factors' you could conjure it up again quickly by asking yourself, "What is six divided by two?" for example.\n\nIn problem (a), you are essentially asking, "What are the coefficients of P(x) if the only factor we know is (x-1)?"\n\nLet's set up synthetic division (see link below), by placing all the coefficients of P(x), and using 1 (the zero implied by (x-1)) as the divisor:\n\n1 ||  a  b  c  d\n___________\n.....a\n\n1 ||  a  b  c  d\n_____a_____\n.....a (a+b)\n\n1 ||  a  b        c       d\n_____a___(a+b)____\n.....a (a+b) (a+b+c)\n\n1 ||  a  b        c                d\n_____a___(a+b)___(a+b+c)_\n.....a (a+b) (a+b+c) (a+b+c+d)\n\nNote that this last term must equal 0 since it is assumed that (x-1) is a factor ... thus, the "remainder" term must equal zero!\n\nThere's the result you were after:  a+b+c+d = 0\n\nHope this helps you tackle the others!
964615
Please type smith's algorithm for polynomial matrices at yahoo search you get links hope fully one of them should provide you with a solution.
142543
Polynomial= multiple terms, i.e. 3x^2+7x-3\nSo I suppose addition of them would be like\n(4x^2+7x-5)+(9x^2+9x+7+y)\n=4x^2+7x-5+9x^2+9x+7+y\n=13x^2+16x+2+y\nor\n(4x^2+7x-5)-(9x^2+9x+7+y)=\n(4x^2+7x-5)-1(9x^2+9x+7+y)=\n(4x^2+7x-5)+ (-1*(9x^2+9x+7+y))=\n(4x^2+7x-5)+(-9x^2-9x-7-y)=\n4x^2+7x-5-9x^2-9x-7-y=\n-5x^2-2x-12-y.\n\nMaybe now you can do your own.\nNote that these methods only work with adding/subtracting.
139386
(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.
293371
The equation has 0 real solutions, but 2 complex solutions.\n\nThere is a very fundamental theorem of algebra which states that any polynomial equation of degree n has n roots. In your case, the degree is 2, so the equation has to have 2 roots.\n\nHowever, roots can be either real, complex, or repeated roots. For example, x^2=0 appears to only have one solutions (x=0) but x=0 is a repeated root, occurring twice. For polynomials of degree 2, you can find out about the solutions by using the discriminant. An equation of the form ax^2+bx+c=0 has discriminant b^2-4ac. The discriminant can be one of three things:\n\n>0 This means that both roots are REAL roots.\n=0 This means that there is ONE REAL root, and it is a repeated root.\n<0 This means that both roots are COMPLEX roots and there is no real solution.\n\nThe methods for general degree n polynomials are much more difficult, if not impossible, to classify.
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
543381
Yes. the RHS of the first is equivalent to xor(P,Q). This becomes an inclusive or when the or(Q,...) is added. Not(P) as suggested above doesn't give the correct result when Q and P both = true.
26326
A procedure for finding the quadratic factors for the complex conjugate roots  of a polynomial P(x) with real coefficients.\n\n[x-(a+ib)][x-(a-ib)]=\nx^2+2ax+(a^2+b^2)=\nx^2+Bx+C.
156051
first let T =X^2\nthan we get a second degree equation\n2T^2-7T-4=0\nso you can solv it easyer
464936
Commutative operation.
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)
902478
wow its been a long time.. i use a calculator to do that now.. but here goes one thing i remember...\n\nlets say we have a polynomial like this\n\nx^2 -3x-10  and we need to factor it..\n\nwe take a look at what factors of -10 we could use that would add up to -3  in this case there are two way to get -10 (-2*5) or (2*-5)  if we add the two sets together one gives us -3 and the other gives us 3 so we can throw out the -2*5 terms.\n\nSo now we lay out the factors.... we need two x's to get x^2 so we need two x's in the first terms (remember FOIL?)\n\nlay the terms out so they give us the correct values\n\n(x+2)(x-5)  now we can use FOIL (firs, outer, inner, last) to check ourselves...\nX*X = x^2\nX*-5 = -5X\n2*x =2x\n2*-5=-10\n\nso we get [x^2] +[-5x] +[2x] +[-10]\nor x^2 -5x+2x-10\ncombine like terms now\n-5x+2x =-3x\n\nwhich gives us\nx^2-3x-10\n\nhope that helps
854837
Show it's a zero just means you sub x = -3 in the polynomial and get the result zero.  If that happens, you can divide the polynomial by \nx -(-3),  i.e. x + 3, and then find the zeros of the remaining expression.  Want to do it yourself?  Good.  Then don't scroll down to see what I've done.\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n-3      2     5     -6    -9\n\n              -6       3      9\n\n         2    -1     -3      0\n\nP(x) = (x + 3)(2x^2 - x - 3)\n       =  (x + 3)(x + 1)(2x - 3)\n\nZeros are -3, -1, 3/2