How Do I Factor (x^4+1)?  Yes...it can be done.?
926495
it can be factored over the complex numbers. \nThe complex roots are:\n(1+i)/sqrt(2), (1-i)/sqrt(2), (-1-i)/sqrt(2), (-1 +i)/sqrt(2)\ni.e.\nx^4+1 = ( x-(1+i)/sqrt(2))( x- (1-i)/sqrt(2)) (x - (-1-i)/sqrt(2) )( x-  (-1 +i)/sqrt(2) )\n'
10610
X^2-9X+18\n\nShort and simple just factor it FOIL
164954
x^3 – 3x^2 + 4x – 12\n= x^2(x – 3) + 4(x – 3)\n= (x-3)(x^2 +4)
178683
yeah, I suppose you want me to factorise x^n - y^n. the general formula is\n\nx^n - y^n = (x-y)(x^n-1.y + x^n-2.y^2 + x^n3.y^3+.........+x^2.y^n-2 + x^n-1.y^1)\n\nTry putting n=2 or 3 or4 in this expression and you'll see what I mean.\n\nYou'll arrive at nothing but\n\nx^2 - y^2 = (x-y)(x+y)\nand so on..\n\nAnd please, the next time, a question without exponents. i'm SICK of '^'. \n\nAnd you know what, mr. 'timmobilelaptop' , the next time try using your brain. It's better than any T-189 calculator.
796585
1.factor\n(2x+1)(x-4)=0\n2.set it to 0\n2x+1=0\nor\nx-4=0\n3.solve\n{x=-1/2 or x=4} <=answer
734537
x^2 +1 has no factors that are real.  I'm not sure whether or not you are into complex numbers yet or not (it has complex factors).\n\nThink about it this way:  It is the graph of x^2 shifted up one unit.  It doesn't touch the x-axis at all, which means it cannot be factored (as the factored form tells you where it touches or crosses the x-axis).\n\nRemember the quadratic formula?  Well, here, b=0, c=1, and a=2, so you end up with b^2-4ac=0-4=-4\n\n-4 has no square root in real numbers.  It's square root using complex numbers is 2i.  "i" is the square root of -1:\nsq(-4) = sq(4) * sq(-1) = 2sq(-1) = 2i\n-b=-0=0, and 2a = 2(1)=2\nThe roots would be +/-2i/2 = +/-i\nThe roots would be thus i and -i.\n\n\nSo the factored form is actually (x+i)(x-i).  Again, this is only relevant if you have learned about complex numbers.\n\n\n\nIf you meant x^2-1, the factors are (x+1)(x-1).\n\nHope this helps.
288929
-> 2x^2 - 1 = 4x + 15\n-> 2x^2 - 1 - 4x - 15 = 0\n-> 2x^2 -4x -16 = 0\n-> 2( x^2 -2x -8 )= 0\n-> x^2 - 2x - 8 = 0\n-> factoring the equation you will come up with\n-> (x - 4)(x + 2) = 0\n-> equate both factors to zero\n-> x - 4 =0 , x + 2 = 0\n-> x = 4, x = -2
394036
Try to solve the eqs with quad rule:\na*x^2 + b*x + c=0\nx=(-b +- Sqrt(b^2-4*a*c))/(2*a)\n\nx^2+2x+5=0 => Get two Complex answers, \nso no it can't be factorized.\n\nx^2+4x+5=0 => Get two Complex answers, \nso no it can't be factorized.\n\nx^2+4x+3=(x+3)*(x+1)=0; \nSo, yes, this can be factorized. \nSolutions are x=-3 and x= -1\n\nx^2+2x-3=(x+3)*(x-1)=0;\nSo, yes, this can be factorized. \nSols are x= -3 and x=1
208014
Factorising gives\n\n(x-4)(x+1) = 0\n\nThe solutions are therefore x = +4 and -1
241036
I'm assuming you're looking for information on factoring polynomials since it's a College Algebra courese.  \nThis website is very helpful, with the steps for factoring laid out, and it also provides websites you can use if you need more information.  Good luck!\n\nhttp://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut7_factor.htm
985516
(8x^4 + x^2 + 9)/x^4\n= 8x^4/x^4 + x^2 / x^4 + 9/x^4\n= 8 + 1/x^2 + 9/x^4 \nor 8 + x^-2 + 9x^-4
888367
You need to give some directions.  What is the problem asking you to do?\nNote: The person below is incorrect.  (4+3x)^5 does NOT equal\n4^5 + (3x)^5
427758
Not factorable unless you use imaginary numbers; if you haven't learned about those, then the answer is no.  \n\nA real solution would be similar to the form\n(3x+5)^2 or (3x-5)^2, but when you square these expressions (FOIL), you end up with an extra term +30x or \n-30x.\n\nAnother method is to apply the . . . can't remember the name, but the formula is this:\n\nFor ax^2 + bx + c = 0, solution may be found by \n\nx = (-b +/- sqrt(b^2-4ac))/2a\n\nfor 9x^2 + 25, b = 0, so when you do the square root part, you will be trying to take the square root of a negative number, which doesn't work for most algebra.  The graph of this function is a parabola that never crosses the x axis and as a result, doesn't have any roots.  Therefore, it is not factorable either.
781910
not possible to factor easily ... try the quadratic formula.
1008674
You got it right so far. But next...to multiply by 1/4...is the same thing as dividing by 4. So take 4x^2/4=x^2 and then -4/4=-1...so you get x^2-1...make sense? I hope I helped...God bless.
119973
(x+3)(x^2-x+4)/(x+3) common factor\n\nmultiply by x+3\n\nx^2-x+4
967957
There really isn't any way to do this other than guess and check. If you find that you substitute some value in and find f(a) = 0, then (x-a) will be a factor. You can repeat that a few times until you have found some roots, and then do a long division to something which you *can* solve by hand (a quadratic or linear).\nNote also that you only need to check things which are factors of the constant term, and there aren't too many.\nIn the first case, you will find that substituting in -2, -1 and 3 will all give 0, but you may have trouble finding the last root. So you will have to do some long divisions (divide by (x+2)(x+1)(x-3)), and you will end up finding that the -1 was a repeated root - so the answer is (x+2)(x+1)^2(x-3).\n\nSame thing for the second: you will find -1 as one root. You can then do a long division and find -1 is repeated, and again, and again, and again. The actual factorisation is (x+1)^5 - you may have noticed this if you knew much about pascals triangle and recognised the coefficients are an entire row - 1,5,10,10,5,1.