factor of x^4+10x^3+26x^2+10x+1?
968937
If it has rational factors they would have to be +-1, neither if which work.\n\nIf there are no linear factors then it must be the product of 2 quadratics:\n\n(x^2 + ax + 1)(x^2 + bx + 1) \n= x^4 + (a + b)x^3 + (ab + 2)x^2 + (a + b)x + 1\n\nso a + b = 10\nab + 2 = 26\n\na = 4, b = 6\n\n(x^2 + 4a + 1)(x^2 + 6a + 1)
817835
x^3-x^2+3x-1\nx+4&#92;x^4+3x^3-x^2+11x-4\n      -(x^4+4x^3)\n              -x^3-x^2\n            -(-x^3-4x^2)\n                       3x^2+11x\n                     -(3x^2+12x)\n                                   -x-4\n                                 -(-x-4)\n                                         0\nx^4+3x^3-x^2+11x-4=(x+4)(x^3-x^2+3x-1)
1157065
(x+5)(2x^2+4x+2)\nx(2x^2+4x+2)+5(2x^2+4x+2)\n2x^3+4x^2+2x+10x^2+20x+10\n2x^3+14x^2+22x+10
1053403
Write it as:\n2x^2-4x+3x-6\n=2x[x-2]+3[x-2]\n=[2x+3][x-2]\nSimple middle-term factorisation!!!
526107
You are only allowed to factor if the equation is equal to zero; your's is equal to 4.  You need to solve "x^3 - 6x^2 +9x - 4 = 0"  using synthetic division.
290738
12x+3-26=6x+1\n\n=> 6x-24=0 => x = 4
83153
There are a couple of ways to work this out, I will take the easiest to explain.\n\n(x-3)(x^3+3x^2+9x+27) Can be rewritten as:\n\n(x)(x^3+3x^2+9x+27)+(-3)(x^3+3x^2+9x+27) Now, multiply those out.\n\nx^4+3x^3+9x^2+27x-3x^3-9x^2-27x-81 Now, sort by like powers of x\n\nx^4+3x^3-3x^3+9x^2-9x^2+27x-27x-81 Simplify\n\nx^4-81
495916
3x^2+9x-84 \nGCF = 3\n3(x^2 + 3x - 28) now we factor the binomial\n\nwe could do this by multiplying the first coefficient (1) by the last coefficient/term (-28). we get -28\nnow we need to select two numbers that multipli to -28 and add to the middle coefficient, 3.\n\nthe two numbers are (7) and (-4)\nwe rewrite the problem with these two terms\n(x^2 + 7x - 4x - 28) i did not write the GCF 3 just yet\nfactor by grouping\n\nx(x + 7) - 4(x + 7)\n\nfactor completely\n\n(x + 7)(x - 4)\n\nfinal answer, don't forget about the 3\n\n3(x + 7)(x - 4)\n\n----------------------------\n4x^2 - 4x - 3   \n1. check for GCF - none\n2. first and last (4)(-3) = -12\n\ntwo numbers that multiply to -12 and add to -4 = 2 & -6\n\n4x^2 + 2x - 6x - 3\n\nfactor by grouping\n\n2x(2x + 1) - 3(2x + 1)\n\n=(2x + 1)(2x - 3)\n\n----------------------------\n3x^2-17x+22 \n\n1. check for GCF - none\n2. first and last   (3)(22) = 66\n3. two numbers are (-6) & (-11)\n4. rewrite\n\n3x^2 - 6x - 11x + 22\n\n5. factor by grouping\n3x(x - 2) - 11(x - 2)\n\n(x - 2)(3x - 11)\n\n----------------------------\nx^2-9 \n\ndifference of squares\n\na^2 - b^2 = (a + b)(a - b)\n\n1. GCF - none\n2. factor\nx^2-9 = (x)^2 - (3)^2 = (x + 3)(x - 3)\n\n---------------------\n\n-3x^2+48?\n\nalso difference of squares\n1. GCF = -3\n\n= -3(x^2 - 16)  \n\n2. factor: x^2 - 16 \nx^2 - 16 = (x)^2 - (4)^2 = (x + 4)(x - 4)\n\n3. final answer = -3(x + 4)(x - 4)
764068
2x - 2 = x^2 - 4x + 3\nSet one side equal to 0\n0 = x^2 - 6x + 5\nFactor\n0 = (x - 5)(x - 1)\nSet each factor equal to 0\n0 = x - 5              0 = x - 1\nSolve\n5 = x                  1 = x
484892
(x+3)(2x+4) is incorrect.\n(x+3)(2x+4) = 2x^2 + 10x + 12\nSolution:\nMultiply the coefficient of the x^2 term with the constant.\n2(12) = 24\nYou are looking for factors of 24 that add the 11 (which is the coefficient of the x term).\n(3)(8) = 24\n3 + 8 = 11\nReplace 11x with 3x + 8x\n2x^2 + 11x + 12\n= 2x^2 + 3x + 8x + 12\nFactor by grouping\n(2x^2 + 3x) + (8x+12)\n= x(2x+3) + 4(2x+3)\n= (2x+3)(x+4)\nAnswer: (2x+3)(x+4)\nYou can easily check the answer by Foil.\n(2x+3)(x+4) = (2x)(x) + (2x)(4) + (3)(x) + (3)(4)\n= 2x^2 + 8x + 3x + 12\n= 2x^2 + 11x + 12
801272
x³y - 3x²y - 10xy --- Take x and y out of each term...\nxy(x² -3x - 10) --- Factor the x² - 3x - 10\nxy(x - 5)(x + 2)\n\n*Note: To show exponents, such as x², use the ^ key (such as x^2). This expression looks like you basiclly wrote:\n3xy - 6xy - 10xy
254834
if you use quadratic formula you get the factors of \n\nx= ( ( 3 +/- square rt.(5) ) / -6 )
532925
9c(4+5c)\nFactors of 12 that add to 7\n(x+3)(x+4)\nFactors of -44 that add to -7\n(x-11)(x+4)\nFirst factor out 5: 5(x^2 + 8x + 16)\nLook for factors of 16 that add to 8\n5(x+4)(x+4)
884668
f(x)=-4x^3+5x^2+8\nf(-3)=-4(-3)^3+5(-3)^2+8\n=108+45+8 is not equal to zero\nso x+3 is not a factor
1174486
First look at the highest powers of a and b which in this case is "squared" and the product 6ab is present, so factoring will likely involve (a+b)^2\nIn fact 3a^2+6ab+3b^2 = 3(a^2+2ab+b^2) = 3(a+b)^2\nSo your problem reduces to:\n\n3(a+b)^2 +7(a+b) - 6\n\nIf we choose to factor -6 as -2 x 3 then the it is easy to factor, and the solution is:\n\n(3(a+b) -2)((a+b)+3)
894442
f(x)=2x^3-3x-1\nf(-1)=-2+3-1=0\nx+1 is a factor\ndividing by x+1\n2  0  -3  -1\n0 -2   2   1\n-------------\n2x^2-2x-1\nso the factors are\n(x+1)(2x^2-2x-1)
913334
x^2 + 2x + 7 Remainder 10\n\ni can add the steps if you don't think i did it by hand