(f^-1) prime (12)   of\n\nf(x)=(1/3)x^3+(5/3)x+2
122316
well, (1/3)x^3 equals x\n\nso you can simplify it to (8/3)x + 2 = f(x)\n\nthat is a good start.
487787
f(x)=16-5x\nf(-3)=16-5(-3)\n=16+15=31
575282
F(n)=F(n-1)+F(n-2), where F(0) = 0 and F(1) = 1\nn= 2 \nF(2) = F(2-1) + F(2-2)\nF(2) = F(1) + F(0) \nF(2) = 1 + 0 = 1\nn = 3\nF(3) = F(3-1) + F(3-2)\nF(3) = F(2) + F(1)\nF(3) = 1 + 1 \nF(3) = 2\nn = 4\nF(4) = F(4-1) + F(4-2)\nF(4) = F(3) + F(2)\nF(4) = 2 + 1\nF(4) = 3\nn = 5\nF(5) = F(4) + F(3)\nF(5) = 3 + 2\nF(5) = 5
259902
I do not know about F.But I know (x).Does she have something like(3,4)(3,5)?\n\nx_|y_ <-- or does she have that?\n3, 4\n\ny=2f(x)\n4=2f(3)                          \n\n      ^\n      That's what she is supposed to do.\n      To solve the function.\n     But still i am sorry about not knowing F.\n    (Those # is example)
673771
f(x) = x^2 + 7x\n\nf(q+2) = (q+2)^2 + 7(q+2)\n\nf(q+2) = q^2 + 4q + 4 + 7q + 14\n\nf(q+2) = q^2 + 11q + 18\n\nf(q+2) = (q + 9)(q + 2)
422316
the prime factorization can be figured out by repeatedly dividing the number by a prime number.\n for example 12\n 12 divided by 2 = 6,   6 divided by 2= 3,  3 divided by 3= 1\n the answer is 1.\ni hope this was helpful.
625828
gurl am not helping out with ur homework next time.\n f(x) =/x/ +3x +1\n       =x+3x+1\n       = 4x+1\n\n  f(x)=3x+/2x-5/\n       = 3x+2x-5\n       =5x-5\n\n  g(x)= /x/+/x-1/\n        = x+x-1\n        =2x-1\n\n  g(x)=3/x-2/ - /x+1/\n        =3(x-2) - x+1\n        =3x-6-x+1\n        =2x-5
439489
c)\n\n= x^3 + x^2 + x - x^2 - x -1\n= x^3 -1
309376
1) 33% (Every third number. 3, 6, 9... 96, 99)\n2) 16% (Every sixth number. 6, 12, 18... 90, 96)\n3) 1% (101 is a prime number, divisible only by 1 and itself.)\n4) 72% (all non-prime numbers)\n5) 28% (1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 49, 53, 59, 61, 67, 71, 73, 79, 83, 89, 91, 97)\n6) 9% (1, 2, 4, 5, 10, 20, 25, 50, 100)
504844
wouldnt it be like 3<x<-3 f(x)\nso like x would be 1 so it would be just f if you really want to simply that far done.
376210
C is correct but not complete... only the 3 is factored out... the real answer would be:\n\n3(x-1)(x+4)
494955
(A) 3+2(x-1)\n\n3+2x+2\n2x+5\n\n(B) 5-2(x+3)\n\n5-2x-6\n-2x-1\n\n(C) 1+2(x-1)\n\n1+2x-2\n2x-1\n\n(D) 6-3(w-2)\n\n6-3w+6\n-3w+12
495347
Having researched this topic as far as possible online, I found very few helpful resources.\n\nWhile the result you mention, using the subsequence 6(5)^n, is intriguing, I suspect that other sequences will yield vastly different bounds on the index.  In particular, is it not true that any prime number p always has a corresponding Fibonacci number F(p+1) of which it is a factor?\n\nGood luck with your own research!
547245
The answer is 2....\n\nUnless you consider 1 as a prime number, which technically it is. Then the answer would be 1
467168
A recap on Remainder and Factor theorem:\nIf f(a) = 0, then (x-a) is a factor of f(x).\n\nWe know that the term in f(x) independent of x is a product of all the roots of f(x), so we can investigate the factors of 9 [+-1, +-3, +-9] in this case to find out the 1st factor of f(x).\n\nBy trial and error, f(3) = 81-90+9 = 0, so (x-3) is a factor of f(x)\nUsing long division f(x) is factorised as follows:\nf(x) = (x - 3)(x^3 + 3x^2 - x - 3)\n\nSo the problem reduces to finding factors for a g(x) = x^3 + 3x^2 - x - 3.\n\nThus similarly, by trial and error, f(1) = 1+3-1-3=0, so (x-1) is a factor of g(x) and therefore f(x) as well.\n\nSo our f(x) becomes f(x) = (x - 3)(x - 1)(x^2 + 4x + 3),\nwhere x^2 + 4x + 3is quadratic and so can be easily factorised to (x + 3)(x + 1).\n\nTherefore, f(x) = (x - 3)(x - 1)(x + 3)(x + 1)\n----------------------------------------------------------------------------------\n\nAlternatively, if you observe that f(x) consists of only even powers of x, you may try to use substitution method to factorise f(x).\n\nLet y = x^2, \n\ntherefore f(x) = x^4 -10x^2 +9 becomes f(y) = y^2 - 10y + 9, which is quadratic and can be easily factorised to give\n\nf(y) = (y - 9)(y - 1)\n\nSubstituting x back, we get\n\nf(x) = (x^2 - 9)(x^2 - 1) = (x + 3)(x - 3)(x + 1)(x - 1), \n\nusing the property a^2 - b^2 = (a + b)(a - b).
584311
(3x-2)(12x^2-2) + (4x^3-x^2+1)(3)\n\nThat is the un-simplified answer
476094
You know that f(x) = 1 + 1/x\nf(1/x) = 1 + 1/(1/x) = 1 + x\nf(x) + f(1/x) = 1 + 1/x + 1 + x\n= 2 + 1/x + x\n\nf(x) * f(1/x) = (1+ 1/x)(1+x)\n= 1 + x + 1/x + 1\n= 2 + 1/x + x\n\nThus, it has been proven.