can lcm(least common multiple) of polynomials be negative? can anyone find the lcm of (1-x) and (x3-1)?
602488
1-x = -(x-1)\nx^3 - 1 = (x-1)(x^2+x+1)\nAnswer: -(x-1)(x^2+x+1)
816611
hi,\n  regarding lcm of a set of numbers.....\n\n  the lowest common multiple (lcm) of two whole numbers\n  is the smallest whole number which is a multiple of both.\n\n  consider the numbers 12 and 15-----\n  the multiples of 12 are : 12, 24, 36, 48, 60, 72, 84, ....\n  the multiples of 15 are : 15, 30, 45, 60, 75, 90, ....\n\n  60 is a common multiple (a multiple of both 12 and 15),  \n  and there are no lower common multiples.\n\n  therefore, the lowest common multiple of 12 and 15 is \n  60.\n\n  now try it yourself for 120 and 130.
529729
(x2)-3 (x-2) over (x2) -4?\nIs this what you mean?\n(x^2)^(-3)*x^(-2)/(x^2)^(-4)\n= x^(-6)*x^(-2)/x^(-8)\n= x^(-6-2-(-8))\n= x^(-8+8)\n= x^0 \n= 1
697928
First you group the 'like terms'; in your example, the first two and the last two. Then you see the common factor and pull it out\n\ny^2(x^2-1) - 15 (x^2-1)\nNow you see the common term in both and pull that out\n(x^2-1) (y^2-15)\n\nthus you have your two factors of the polynomial
745882
C is correct but not complete... only the 3 is factored out... the real answer would be:\n\n3(x-1)(x+4)
150854
i bet your thinking, my god, it has 3 values (x,y,z), how the heck does one attack this\n\nbelieve it or not, its simple\n\njust like you would in R^2\n\nsqrt[ (x-x1)^2 + (y-y1)^2 + (z-z1)^2]\n\nso \n\nsqrt [ (3-1)^2 + (-3- -4)^2 + (2-7)^2]\n\nso\n\nsqrt[ (2)^2 + 1^2 + (-5)^2]\n\nso\n\nsqrt [ 4 +1+25]\n\nso\n\nsqrt [30]\n\nhope that helps
51216
x = [-(-4) +/- root((-4)^2 - 4x1x(-5))]/2x1\n  = (4 +/- 6)/2\n  = 5 or -1
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.
333704
1. False, a polynomial has no = sign.\n2. False, the log of zero (or any negative number) does not exist.
920343
(4/3)[x^3/(2-x)]^(1/3)\n[(2-x)(3)x^2-x^3(-1)]/\n(2-x)^2=\n\n 4x[1/(2-x)]^(1/3)\n[3x^2(2-x)+x^3]/\n3(2-x)^2
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
915401
this means\n\nx^3 = -1\nor x^3+1 = 0\n\nf(x) = x^3+1\n\nf(-1) = 0\nso one root is -1\n\nfor the other root by division\n(x+1)(x^2-x+1)\nx^2-x+1 = 0 => x= (1+/-sqrt(3)i)/2
652975
(x+4)/[(3x-1)(3x+1)]-\n(x-2)/(3x+1)=\n\n[(x+4)-(3x-1)(x-2)]/\n[(3x-1)(3x+1)]=\n\n[x+4-3x^2+6x+x-2]/\n[9x^2-1]=\n\n[-3x^2+8x+2]/[9x^2-1]
599219
you have to solve the ones in the brackets first. so in this instance, 4(-1) = -4.\n\nand then, you solve -4 - 3 = -7
818533
For LCM, keep dividing by common prime factor, starting from 2, throughout till you get 1 in the end.\nInitially 6, 15, 20\nDivide by 2: 3, 15, 10 (You can't divide 15 by 2, hence you leave it as it is.)\nDivide by 2: 3, 15, 5 (Similarly, you can't divide 3 by 2, hence you leave it as it is.)\nDivide by 3: 1, 5, 5\nDivide by 5: 1, 1, 1\nLCM = 2 x 2 x 3 x 5 = 60\n\nHope this helps :)
593998
When you have a polynomial, the coefficients are integers.  For example:\n\nax^2 + bx + c = 0\n\nThis is the general formula for a second degree polynomial (x is to the 2, hence second degree).  In this case, a, b, and c are the integers.\n\nAlso, if you add other terms, such as x^3, they would also be multiplied by integers.
173410
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