Factor 1/9x^2+4x+36, please?
444843
It's a perfect square trinomial.  Take the square root of the first term and the square root of the last.  If you multiply them and then double it, and it equals the middle term, then it's a perfect square trinomial.\n\nIn this case, it factors into (x/3 + 6)^2.  Using FOIL, you'll see that it expands into your original expression.
800533
Well dear Kristi \n\nthe answer is \n( 9r - 2s )^2 \n{ as you know ( a - b )^2 = a^2 -2ab+ b^2 \n81r^2 - 36rs + 4s^2\na = 9r , b = 2s , so \na^2 = (9r)^2 = 81r^2 \nb^2 = (2s)^2 = 4s^2 \n-2ab = (-2)(9r)(2s) = -36rs  } \n\nGood Luck Darling
284003
(x-6)/4 = (x-9)/2\n\nCross-Multiply\n\n2(x-6)=4(x-9)\n\nSolve\n\n2x-12=4x-36\n-12=2x-36\n24=2x\n\nANSWER : x=12
955059
I think it's 9*4 - (9+4) = 36 - 13 = 23.
930308
x=[2+/-rt(4+60)]/2\n=[2+/-8]/2\n=-3 or 5\nso the factors are\n(x+3)(x-5)
478149
-4 (4r^2 + 9) +9 (8r + 2)\nDistributive Propert, (-4)(4r^2)+(-4)(9)+(9)(8r)+(9)(2)\nSimplify, -16r^2-36+72r+18\nCombine like terms, -16r^2+72r-18
954091
-4t^2+36-72\nFactor-2t+12\n +2t-4t^2+24\n-6+12-72\nIt is (2t-6)(-2t+12)\n\nFirst you put the first trinomial in the upper left corner of the box (-4t^2).\nThen you put the last trinomial in the bottom right corner of the box(-72).\nYou multiple (-4t^2 & -72) then you make a list of factors that will equal that number. In this case they are 12 & 24.\nYou know that one number must be negative because -4t^2. Fill in the info in the box and multiple them out. \nIf you do this you can see that it equals the trinomial.\nAdd down the left side of the box to get the first set of numbers.\nThen add across the box to get the second of numbers.\nYou can see how I did this.\n\nYou can email me if this doesn't make sense.
930643
factor:\n\n(2x^2+x-6)/(2x+4)=\n\n(2x-3)(x+2) / 2(x+2)\n\nthe (x+2)s cancel out leaving\n\n(2x-3)/2
651993
Here's the best I can do to help with the first one.\n   __x_+9___\nx-2)X^2+7x-2\n    x^2-2x\n    ------\n        9x-2\n        9x-18\n        -----\n           16\nFollow the example for the rest
833702
Do you mean (2x^2 + 3x + 1/ 2x^2 + x - 3) for your first fraction?  I'm going to assume you do.\n\nFirst, factor each numerator and denominator:\n\n2x^2 + 3x + 1 = (2x+1)(x+1)\n\n2x^2 + x - 3 = (2x-3)(x+1)\n\n4x - 4 = 2(2x-2)\n\nNow let's look out our multiplied fraction, rewritten using our factors we just found:\n\n[ (2x+1)(x+1)2(2x-2) ] / [ (2x-3)(x+1)(2x+1) ]\n\nNotice we can cancel out the (2x+1)'s and the (x+1)'s, to get:\n\n2(2x-2)/2x-3\n\nANSWER:   2(2x-2)/2x-3
216863
((4 * 4) / 4) - 4 + 4 = 4
958614
#1.)-----------------------------------\n  \n25b^ -20by + 4y^\n\nMultiply 1st and 3rd term together to get 100b^y^.\n\nFactor 100b^y^ into two factors whose sum = the middle term of -20by.\n\n1by   X 100by\n2by   X 50by\n4by   X 25by\n5by   X 20by\n10by X 10 by   ....................BINGO!\n\n-10by + (-10by) = - 20by   &   (-10by)(-10by) = 100by\n\nThis is a really kewl factor by grouping trick, I LOVE it!\n\nBring down the first term of: (25b^)\n\nNext, write the SMALLER of the two factors 1st.  Followed by the LARGER of the two factors.  In this case, they're both the same, so it doesn't matter which one goes first: (-10by)\n\nFinally, bring down the third term of: (+4y^)\n\n25b^ -10by -10by + 4y^\n\nNow, lets factor by grouping! \nFactor out the GCFs.\nGCF#1 = 5b\nGCF#2 = -2y\n\n(25b^ -10by) + (-10by +4y^)\n5b(5b - 2y) -2y(5b - 2y)\n\nGCF#3 = (5b - 2y)\n\n(5b - 2y)(5b - 2y)\n\nSimplifies further as: (5b - 2y)^   ....Whatdayaknow?! \n.......................................................................................................It's a Perfect Square Trinomial.\n\n#2.)------------------------------------------------------------------\n\n100 - 9b^\n\nAh, the difference of two perfect squares.\n\n---------------------------------------------\n\nFORMULA:\na^ - b^ = (a-b)(a+b)\n\n---------------------------------\n\nSOLUTION:\n100 - 9b^\n(10)^ - (3b)^\n\n-----------------------\n\nANSWER:\n(10-3b)(10+3b)
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
517819
start pulling out common factors\n\n2 (x^3 + 27)\n\n2(x^3 + 3^3)
494989
It seems that many people here don't realize that your equation is 3x^2 - 6x + 9.\n\n3 is the basic factor to get 3*(x^2 - 2x + 3)\n\nIf you are allowed to use imaginary numbers (where i = sqrt(-1)) then,\n\n3*[x - 1 + i*sqrt(2)][x - 1 - i*sqrt(2)]
1059114
3x^2 - 9x + 6 - 3(x - 1)(x - 2)\n3x^2 - 9x + 6 - 3(x^2 - 2x - x + 2)\n3x^2 - 9x + 6 - 3(x^2 - 3x + 2)\n3x^2 - 9x + 6 - 3x^2 + 9x - 6\n\nEverything gets cancelled out, so it doesn't matter if i put 10 in for x or not.\n\n3(10)^2 - 9(10) + 6\n3(100) - 90 + 6\n300 - 90 + 6\n210 + 6\n216
534750
1/2 + 1/2b + 2b^2 + 6bc + c^2\n\n(1/4b)(1/4b) + 4bc + (bc)(bc)\n1/2 + 1/4b +1/4b + b^2 +4bc + b^2 + c^2 + 2bc\n1/2 + 1/2 + 2b^2 + 6bc + c^2\n\nI think this is right... email me if I'm wrong?