Can u plz factor 12-13b+b^2?
397605
12-13b+b^2=(12-b)-(12b-b^2)\n  = (12-b)-(12-b)b\n  =(12-b)(1-b)\n\nThis might not be easy to see, so here is the general approach:\n  Cb^2 + Db + E = (b-B1)(b-B2)\nwhere B1 = (-D + sqr(D^2-4CE) ) /(2C)\n and B2 = (-D - sqr(...) ) /(2C)\n\nIf (D^2-4CE)<0, you cannot factor it.\n\nsqr is square root. In your case C=1, D=-13, E=12, so you get B1=12 and B2=1
595135
88 x 132 = (87+1)x132=87x132 + 1 x132=N +132\n87 x 132= N-0
631080
You can noy factor it out... it makes it easier to figure out when you put it into lowest terms... so it would be \n\n   x^2+7x-25..... so you cant factor it.
742127
= (x^12)/12 - (8x^11)/11 + (5x^10)/2 - (10x^9)/3 - (15x^8)/4 + 24x^7 - 49x^6 + 60x^5 - (195x^4)/4 + (80x^3)/3 - (19x^2)/2 + 2x + C
945505
{(x^2+5x+4)/(x+3)^2}/(x+1/(x^2 -9))\nFactor everything that you can factor.\nx^2 + 5x + 4 = (x+1)(x+4)\nx^2 - 9 = (x+3)(x-3)\nYou have\n{(x+1)(x+4)/(x+3)^2} * {(x+3)(x-3)/(x+1)}\nYou can cancel the (x+3)'s and the (x+1)'s leaving you with\n(x+4)(x-3)/(x+3)
113395
8 = 2*2*2\n12 = 2*2*3\n52 = 2*2*13\n\nSo it's 2*2 = 4
777394
factor out 4b\n\n4b(b-2)= 0\n\nb = 2, 0\n\ngoodluck ^_^
821433
$13212 = $(12000 + 1200 + 12)
779652
Do you mean a right triangle, if so the answer is 13. 5^2 + 12^2 = 13^2. If not, it depent on the angles.
738298
[k+2]=13\nk+2-2=13-2\n       k=11
977376
at the first 4 you use a^3-b^3=(a-b)(a^2+ab+b^2)\n and a^3+b^3=(a+b)(a^2-ab+b^2)\n\nfor the next 2 :\nthe first 2 factors and the second 2 factors have something in comon\n\nthe next ones use the above formulas plus one more \na^2-b^2=(a-b)(a+b)
944695
12 * (3-e^-3x)^3 *  e^-3x
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)
81890
it's not factored completely if you leave it as a^2-25 so the proper answer is 5(a-5)(a+5)
165354
To factor x^2 - 100 by the difference of 2 squares, figure out what times itself = x^2, and what times itself = 100\n\nThen put the two answers (x and 10) in binomials like this:\n\n(x+10)(x-10)\n\nWhen you multiply this out by FOIL (at least that's what we call it), the -10x and the 10x "cancel out" and all that's left is x^2 - 100
517819
start pulling out common factors\n\n2 (x^3 + 27)\n\n2(x^3 + 3^3)
259003
Use the quadratic formula.  It works everytime!\nx = (-b plus or minus sprt(b^2 - 4ac))/2\nwhere\nax^2 + bx + c = 0\nSome of these will factor, however not all of them will.