what is the commnon factor of x+3, 3x, and 2x+6?
916165
Sorry, there isn't one.  If you meant "3x + 9" instead of 3x, then x + 3 is a common factor.
825419
x^3 - 6x^2 - 3x + 18 \n= x^2 (x -6) - 3(x-6) \n= (x^2 - 3)(x - 6)
798514
do you mean 6x([1/2]x^2)?\nif so 3x^3\nor 6x(1/(2x^2))?\nthen its 3x^-1
234504
Multiply 3 times each of the factors in the parenthesis\n3x+12=27\nmove all like factors to each side, rememberr to switch signs if you move a factor to the other side\n\n3x=27-12\n\n3x=15\n divide each factor by 3\nx=5\n\nX is the variable
21241
=2x(x+6)+2(x+6)\n=2x^2+12x+2x+12\n=2x^2+14x+12\n\nRefer the below site for the concepts. Its quite nice.
869535
5x+8+3x-5=6x-3\n8x+3=6x-3 (combine like terms)\n2x=-6 (subtract 6x, subtract 3)\nx=-3 (divide by 2)
1046507
[2x/(x+3)(x-3)]*[(x-3)/4(x)]\ncancelling 2 and x-3 we will get\n=x/(x+3)2(x)\nsimplify we will get\n=x/(2x^2+6x)
541499
2x(x-5)-3(2x+1)\n2x^2-10x-6x-3\n2x^2-16x-3
829616
First, factor the numerator:\nx + 2x² + 3x - 6 --- Combine like terms...\n2x² + 4x - 6 = (2x - 2)(x + 3) = 2(x -1)(x + 3)\n\nNow, you can cancel out the (x - 1) from the numerater and the denominator...\n2(x + 3) / 1 = 2(x + 3)\n\n2(x + 3) is the answer...
905618
(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
339354
(2x + b) * (a + 3c)\n\n\nsteps:\n2ax + 6xc + ba + 3bc \n1) group first two, and last two\n(2ax + 6xc) + (ba + 3bc)\n\n2) factor out 2x from the first, and then b from the second group\n2x*(a + 3c) +  b*(a + 3c)\n\n3) since (a + 3c) is common, combine 2x and b\n(2x + b)*(a + 3c)
223322
3x*(x^2-8xy+16y^2) Same factor\n3x*(x-4y)^2 Perfect squares.
222653
x+6(x^2)= 12\nx+6x^2  = 12\n6x^2+x-12=0\n(2x+3)*(3x-4)=0\n2x+3=0 3x-4=0\n2x=-3  3x=4\nx=-3/2 x=4/3
503348
No, the previous answer is wrong.\n\nGroup it like this:\n\n2p + pq -6q -3q^2\n\nthen factor like this:\np(2+q) -3q(2+q)\n\nfactor out the 2+q to get:\n(2+q)(p-3q)\n\nFOIL it out and you will see it is correct.
159156
factoring, then:\n\n(2x-3)(x+1)=0\n\nyou can assume that\n\n2x-3 = 0, and\n\nx+1 = 0\n\nSolve for first x,\n\n2x=3 , x=3/2\n\nsecond x,\nx= -1\n\nThe sum would be 3/2 - 1, or 3/2 - 2/2 =1/2
335000
(3x-9)/(x^2-x-6)=\n\n[3(x-3)]/[x^2-3x+2x-6]=\n\n[(3(x-3)]/[x(x-3)+2(x-3)]=\n\n[3(x-3)]/[(x-3)(x+2)]=\n\n3/[x+2]
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