factors of x^4+6x^3+17x^2+36x+66?
962913
because it is quartic equation it can be factored into 2 quadratic with real coefficients. let us try if real\nx^2+ax+b and x^2+cx+d as coefficient of x^4 is 1\n\nmultiply and equate coefficent on both sides\n\na+c = 6 (x^3)\nbd = 66(constant)\nb+d+ac = 17(x^2)\nad+bc = 36(x)\n\nif they are rational they are integer\nfrom bd =66 let b= 6 and d= 11(we can chose one)\nthen a+c = 6\nac =0\n11a+6c = 36\nso a= 0 and c = 6\nequation is consistat\nso we get (x^2+6)(x^2+6x+11)\n\nthese cannot be factored without going to irrational and complex
36762
Factor by grouping:\nx^2 (3x - 5) - 3 (3x - 5)\nAnswer: (3x - 5)(x^2 - 3)
950667
This is a mathematics question, but fine. \nThe factors are 1, 2, 3, 4, 6, 8, 12, 16, 24.\nThe prime factors are 2^4,3.\n\nJust trial and error.
547775
f(x)= x^3+x^2-22x-40\nf(x)= (x+4)(ax^2 + bx + c)\n\nComparing coefficients,\nx^3: 1=a\nconstant: -40=4c => c=-10\nx^2: 1=b+4a => 1=b+4 => b=-3\n\nf(x) = (x+4)(x^2-3x-10) = (x+4)(x+2)(x-5)\n\nThe other factors are (x+2) and (x-5).
821200
(x²+4x+4) / (2x-3) / (x²-4) / (2x²-7x+6)\n\nYou have 4 terms and three divisions, so it's a little unclear if you mean (a/b) / (c/d) or what.  I will try to help to the extent I can.\n\nLet's assume you mean:\n\n[ (x²+4x+4) / (2x-3) ] / [ (x²-4) / (2x²-7x+6) ]\n\nFactor your x² equations to see if you can cancel terms:\n\nx²+4x+4 = (x+2)(x+2)\nx²-4 = (x+2)(x-2)\n2x²-7x+6 = (2x-3)(x-2)\n\nWoohoo - things are looking good!  Now we can rewrite our original problem using these factors:\n\n[ (x²+4x+4) / (2x-3) ] / [ (x²-4) / (2x²-7x+6) ] = \n\n[ (x+2)(x+2) / (2x-3) ] / [ (x+2)(x-2) / (2x-3)(x-2) ] =\n \n[ (x+2)(x+2) / (2x-3) ] * [ (2x-3)(x-2) / (x+2)(x-2) ] =\n\nnotice we can cancel out three terms: (2x-3), (x-2) and (x+2)\n\n[ (x+2)(x+2) / (2x-3) ] * [ (2x-3)(x-2) / (x+2)(x-2) ] =\n\n(x+2)\n\nHope that helped!\n\n~ ♥ ~
754111
devide by 3\n\nx^4+2x^3-41x^2-42x + 360 = 0\nnow if has a real solution it is factore of 360.\n1/-1 2/-2 can be tried out\n\nf(3) is zero so x-3 is a factor\nx^4-3x^3+5x^3-15x^2-26x^2+78x - 120 x + 360 =0\n\nor(x-3)(x^3+5x^2-26x-120) = 0\n\nx = 3\nor say g(x) x^3+5x^2-26x-120 =0\n\ng(5) = 125+125-130-120 = 0\nso x(-5) is a factor\n\ng(x) = (x-5)(x^2+10x+24) \n\nx= 5 or x^2+10x+24 =0\n\nx = -6 or -4\n\nso x = 3 or -4 or 5 or -6
810321
you will need to use the quadratic formula since you cannot easily find factors\nfrom the given equation above a=1  b= -7  c=7 (where a,b,c are the co-effiecients of the given eqn respectively)\n\nusing the quadratic formula x= -b(plus or minus) under the square root sign(bsquared- 4ac) all divided by 2a\n\nthen x= -(-7) (plus or minus) under square root sign (-7)(-7) - 4 (1*7) divided by 2(1)\n\nx= 7+ or - square root 49-(4*7) /2\n\nx=7+ root of 49-28 all divided by 2\n\nx=7 +or - root of 21 all divided by 2\n\nx=7+ 4.58 divided by 2\nthen x=5.79=5.8\nor\nx=7-4.58 divided by 2\nx=1.21=1.2\ntherefore\nx=1.2, x=1.8\ni ope u understand wat i ve done\nthe needed symbols and signs arent availabe that is y i wrote sum of the words out\nall the best
652953
I got:\n1. 6x^2 - 36x +48\n2. 6x + 22
204966
I can solve it if you'll clarify whether the equation is\n2x^2 = 17/3x + 1 OR 2x^2 = (17/3)x + 1 OR 2x^2 = 17/(3x+1).\nI'm not posting a bunch of steps for three different equations.\n\nI'll post this one (it's the easiest to explain):\n\n2x^2 = (17/3)x + 1\n\nMove all the terms to left side of the equal sign.\n2x^2 - (17/3)x - 1 = 0\n\nClear the fraction by multiplying all terms by 3.\n3(2x^2) - 3(17/3)x - 3(1) = 3(0)\n\nThe equation is now:\n6x^2 - 17x - 3 = 0\n\nNow you multiply the coefficient of the first term by the third term:\n(6)(-3) = -18\n\nNow you need two factors of -18 that when added equal -17.\nIn this case, they are -18 & 1.\nSubstitute them for the middle term.\n6x^2 + (-18 + 1)x - 3 = 0\n\nRemove the parentheses and distribute the x. The equation becomes:\n6x^2 - 18x + x - 3 = 0\n\nExtract the greatest common factor for the first two terms:\n6x(x - 3) + x - 3 = 0\n\nNow extract the greatest common factor for the last two terms:\n6x(x - 3) + 1(x - 3) = 0\n\nRewrite the equation as its factors:\n(6x + 1)(x - 3) = 0\n\nSet the value of each factor to 0:\n6x + 1 = 0 & x - 3 = 0\n\nSolve each:\n\n6x + 1 = 0\n6x = -1\n6x/6 = -1/6\nx = -1/6\n\nx - 3 = 0\nx = 3\n\nSo, x = -1/6 & x = 3
300260
Alright, I've got two answers using the four basic operations:\n\n(14 - 6) * 3\n\nand\n\n(6/(1-(3/4)))\n\nIf you allow other operations, there are a few more answers (! is the factorial symbol : 4! = 4*3*2*1, 3! = 3*2*1, etc)\n\n4*1*sqrt(36)\n(3*6-14)!\n16*3-4!
528350
(5x + 4)(6x + 4)\n30x^2 + 20x + 24x + 16\n30x^2 + 44x + 16\nANS : 16
1070097
(x+3) (5x-3) (7x-1)\n\nYou can do this manually by figuring out the factors of 35 and 9.\n\nFor 35:  {1, 5, 7, 35, -1, -5, -7, -35)\nFor 9: {1, 3, 9, -1, -3, -9)\n\nI started assuming that (x + n) would be a factor.  I tried (x + 1), (x - 1), then (x + 3).\n\nIf you do long division, you get:\n(x + 3) ( 35x^2 - 26x^2 + 3)\n\nThen you just repeat the process for ( 35x^2 - 26x^2 + 3)... you need factors of 35 and factors of 3 that will add up to -26.  \n-3 x 7 = -21 and -1 x 5 = -5, so (5x - 3)(7x - 1) works.\n\nThat gives the final answer of: \n(x+3) (5x-3) (7x-1)
485469
product of 8 and -3  = -24\nfactors for -24 are 1)-12 ,2\n                    2) 12, -2\n                    3) 4 , -6\n                    4) -4 , 6\n                    5) 8, -3\n                    6) 3, -8\n\nOf the above only the factors 1), that is -12, 2 when added give -10.\n\nTherefore we would take those factors. And split the middle term to get the following:\n\n               -3x^2 -12x + 2x + 8\n\nNow factorize to get:   -3x( x + 4 ) + 2 ( x + 4 )\n                      = (-3x + 2)( x + 4)
433699
factor the denominator:  x^2 - 4x - 45 = (x+5)(x-9)\n\nSo, (x+5)/(x^2 - 4x - 45)=(x+5)/[(x+5)(x-9)]=1/(x-9)
640891
I'm guessing you meant 17x^3 - 10x^2 - 6x +4 and 2x^2- 6x +2\nIf that's right, you just add like powers of x and their coefficients to get 17x^3 - 8x^2 -12x +6.
1095294
5) find the factors\nsquareroot of x * x * x * x * x * x /  7*7*5*5*2*2\nthe factors can come out of the square root by pairs only\nyou have 3 pairs of x's = x^3\nyou have 1 pair of 7's so it comes out of the square root as a 7\nx^3/7*5*2\nx^3/70 is it's simplified term\n\n__________________\nif u want all the correct answers then that would tarenirator's answers
413378
24, factors are 1, 2, 3, 4, 6, 8, 12, 24