4x+8 divided by 6x-3x^2 multplied by 3x^2-4x-4 divided by 9x^2-4
130462
You need to factor everything, then cancel common factors.\n4x+8 factors as 4(x+2)\n6x-3x^2 factors as 3(2-x)\n3x^2-4x-4 factors as (3x+2)(x-2)\n9x^2-4 factors as (3x+2)(3x-2)\n\nCancel the (3x+2)'s, and the (x-2) can cancel with the (2-x) leaving a factor of -1.  The factors that are left are:\n[-4(x+2)]/[3(3x-2)].
231686
It has been a long time, here are suggestions:\n1. 5x^3 + 11x^2 + 3x + 1  \n2. 4x^3 - 13x^2 - 8x + 7 \n3. 2x^4 - x^3\n4. 21x^2 + 26 x + 8
3086
(([(4+4+4).(4+4)]+4)/4)*(4+4)\n\n(4+(4*4)).([(4+4+4).(4+4)]+4)
646574
(2x-9)(2x+9)\n\na guess... its been a long time since I did that stuff
511928
8x-10x > -11-1\n\n-2x > -12\n\nx < 6
333631
(3x + 6/4(x^2)) (x + 2/16(x))\ntake out x\n=x(3 + 3/2(x)).x(1 + 1/8)\n=9/8(x^2)(3 + 3/2(x))\ncan simplify d numbers still.
460869
1) 4-3x<_8+x\n4+8<_x+3x\n12<_4x\n3<_x or x in the interval [3,plus infinity)\n\n2) 3x-3>-x+1\n3x+x>1+3\n4x>4\nx>1 or x in the interval (1, plus infinity)\n\n3) 3(x-1)_<4x\n3x-3<_4x\n-3<_x or x in the interval [-3, plus infinity)\n\n4)1/2x<x+1\n-1<x/2\n-2<x or x in the interval (-2, plus infinity)\n\n5)-1/2x>1-x\nx/2>1\nx>2 or x in the interval (2, plus infinity)
487385
2x^2-8x+3x-12\n2x(x-4)+3(x-4)\n2x+3 = 0 or x-4= 0\nx = -3/2 or x = 4\n\n2x^2+10x-72\n2x^2+18x-8x-72\n2x(x+9)-8(x+9)\nx = 4 or x = -9\n\nx^2+6x-x-6\nx-1 = 0 or x+6 = 0\nx = 1 or x = -6
499022
12x^2-10x-8\n=2(6x^2-5x-4)\nsplitting the middle term -5 into\n-8 and +3 so that the sum is -5 and\nthe product is -24\n=2(6x^2-8x+3x-4)\n=2(2x+1)(3x-4)
459235
9c^2 - 64\n(3c+8)(3c-8)\n\n9c^2-24c+24c-64\n9c^2-64
93326
You've just got to do the cross-multiply trick.  Sooo...\n4(x-2)=24 -> 4x-8=24 -> 4x=32 -> x=8
439647
Try factoring by grouping, the first 2 then the last 2.\n\n(x^4+x^3)(-27x-27)
470251
[(x1+x2)/2, (y1+y2)/2]\n(1,8)
482020
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
96179
x^4 + 4\n\nWell, you can add 0 in the form of +a - a, so\nx^4 + 4 +4x^2 - 4x^2\n\nRegrouping them, we get\nx^4 + 4x^2 + 4 - 4x^2\n\nUsing factoring, we get\n(x^2 + 2)^2 - (2x)^2\n\nThen,\n[x^2 + 2 + 2x][x^2 + 2 - 2x]\n\nRearranging them for a more beautiful sight, we get\n{x^2 + 2x + 2)(x^2 - 2x + 2)\n\nThanks For the Problem ^_^
415430
(x+3)(x^2+2x+4)\n\nStart by taking the first term in the first parenthesis and multiplying it by each term in the second parenthesis:\n\nx*x^2 + x*2x + x*4 = x^3 + 2x^2 + 4x\n\nNow take the second term and multiply it by each term in the second parenthesis:\n\n3*x^2 + 3*2x + 3*4 = 3x^2 + 6x +12\n\nNow add all of that together\n\nx^3 + 2x^2 + 4x + 3x^2 + 6x +12\n\nand combine terms\n\nx^3 + 5x^2 + 10x +12
133880
the first part is the same as 8+(14-20)x = 8-6x\nthe second part is the same as 5+2x-2 = 3+2x