I need help solving/factoring these problems?
967876
1) 2a^3 - 13a^2b + 11ab^2 + 14b^3\n\n2) (2a-4b) (4a^2+8ab+16b^2)\n\n3)  (b-8)(b-1)=0\n      b=8   or    b=1\n\n4) 4m^3 - 26m^2 - 14m=0\n2m(2m^2 - 13m - 7)=0\n2m=0  or  2m^2 - 13m -7=0\nm=0  or    (2m+1)(m-7)=0\n                  m=-0.5   or  m=7\n\n so the final answer is m=0 and m=-0.5 and m=7
821200
(x²+4x+4) / (2x-3) / (x²-4) / (2x²-7x+6)\n\nYou have 4 terms and three divisions, so it's a little unclear if you mean (a/b) / (c/d) or what.  I will try to help to the extent I can.\n\nLet's assume you mean:\n\n[ (x²+4x+4) / (2x-3) ] / [ (x²-4) / (2x²-7x+6) ]\n\nFactor your x² equations to see if you can cancel terms:\n\nx²+4x+4 = (x+2)(x+2)\nx²-4 = (x+2)(x-2)\n2x²-7x+6 = (2x-3)(x-2)\n\nWoohoo - things are looking good!  Now we can rewrite our original problem using these factors:\n\n[ (x²+4x+4) / (2x-3) ] / [ (x²-4) / (2x²-7x+6) ] = \n\n[ (x+2)(x+2) / (2x-3) ] / [ (x+2)(x-2) / (2x-3)(x-2) ] =\n \n[ (x+2)(x+2) / (2x-3) ] * [ (2x-3)(x-2) / (x+2)(x-2) ] =\n\nnotice we can cancel out three terms: (2x-3), (x-2) and (x+2)\n\n[ (x+2)(x+2) / (2x-3) ] * [ (2x-3)(x-2) / (x+2)(x-2) ] =\n\n(x+2)\n\nHope that helped!\n\n~ ♥ ~
638676
1.)\nstep one: if they are zeroes, use opp. \nexample....\n (x-2) (x+9) (x-7)          if you solve for x it should be opposite of given zeroes.\n\nstep two: FOIL the first two...\nyou'll get   x^ +9x -2x -18 which simplifies to\n\n(x^2 + 7x -18) (x-7) \n\nstep three: FOIL these two together...\nyou'll get     x^3 +7x^2 - 18x -7x^2 -49x + 126\n\nwhich simplifies to  \n   x^3 - 67x +126\n\n\n2.) I don't know about the second one... I could solve it if I spent some time, but start with...\n (x - 4i) (x+5) (x + 2i)\nthan just do what I did in the problem above, factor the first two, than factor your answer with the last one... and remember that i^2 is equal to 1...
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
134530
First, generate 3 equations by inserting the values of x and y for each of the coordinates.\n\nEqn 1. For (-1,1): 1=a*-1^2+b*-1+c yields: 1=a-b+c\nEqn 2. For (1,3): 3=a*1^2+b*1+c yields: 3=a+b+c\nEqn 3. For (3,2): 2=a*3^2+b*3+c yields: 2=9a+3b+c\n\nStep 1: solve Eqn 1. for a:  a=1+b-c\nStep 2: insert a from Step 1 into Eqn 2.: 3=(1+b-c)+b+c ==> 2=2b ==> b=1\nStep 3: insert b=1 back into Step 1 and re-solve for a: a=1+1-c ==> a=2-c\nStep 4: insert the results from Step 2 for b and Step 3 for a into Eqn 3 to find c.\n  2=9*(2-c)+3*1+c ==> 2=18-9c+3+c ==> 19=8c ==> c=19/8\nStep 5. Insert the results from Step 4 and Step 2 into Eqn 1 to find a.\n  1=a-1+19/8 ==> a=-3/8\n\nTherefore:\n  a=-3/8\n  b=1\n  c=19/8\n\nStep 6: Verify for -3/8*x^2 + 1*x +19/8\n (-1,1): 1=-3/8*(-1^2)+1*-1+19/8\n (1,3): 3=-3/8*(1^2)+1*1+19/8\n (3,2): 2=-3/8*(3^2)+1*3+19/8
1020883
i am sure you can find a math tutor that will be happy to help you when you explain what the problem is.\n\nso what IS the problem, anyway?
1129053
Go to the site below.  It works great, you just put the colors where they belong and click solve, it gives you a step by step on how to solve it.
195201
i think u typed the question wrong, can u please correct it and if u have 2 unknowns, please give 2 equations, else u cant get a definite solution
836789
All this problems are solved using the equation:\n\nHeat = M * Cp * DeltaT\n\nPlug the numbers, check wich one is OK.  Oh, and watch the units.
571026
Whenever you have the difference of two squares\n\nF^2-S^2\n\nit factors as\n\n(F-S)(F+S)\n\nMemorize that pattern and problems like this will be a breeze.
898860
First one: d= - 1\n\nEDIT: the rest I'm too lazy to do. Do your own homework! =D
953940
Hi again \n\nY = √2x – 3 \n2x – 3 = y ²\nX =( y ² + 3 )/ 2\nf –¹ (x)  =( y ² + 3 )/ 2  =>   f –¹ (x)  = (x ² + 3 )/ 2   &  x ≥ 0 \n\nGood Luck \n try to solve them yourslef again if you got wrong , am sure you can solve them correctly , if you just try your best
68882
1) yes, this is correct as written.  Not sure what else you need? \n2) 10/14?\n3) 2 (1 is not a factor)\n4) -6(x+1)\n5) 5 1/4 or 21/4 (same thing)\n6) 1 4/9
148434
So, continuing with mr_w's excellent example, we have 4 equations in 4 unknowns. So you can set this up as a matrix-vector problem.  Define matrix M such that M * Number of games bought vector = Cost & total games vector.  Assigning the rows of M in order to the total cost, the PS3 and Xbox360 cost, the PS1 and Xbox360 cost, and the total number of games, \nM = \n70 150 300 450\n0 0 300 450\n70 0 0 450\n1 1 1 1\nNumber of games bought vector =\na\nb\nc\nd\nand Cost & total games vector =\n13910 12000 3050 54\nThe solution will find the values of a, b, c and d.\nWhen I ran the solution I ended up with noninteger values for each and the values checked with the problem statements.  I hope those values were mr_w's solution.
255329
Have patience :)  You got some answers the first time you asked this.  Here it is again:\n\nIt does not get eliminated, it gets split up into the 2 factors.\n\nOn this one, you can go ahead and divide by 2 to make things a little easier. You'll be left with\n\n3x^2 - 7x - 6 = 0\n\nNow to factor. You need 2 numbers whose product equals 3*-6, or -18 and whose sum equals -7. From this point on it's just sort of educated guessing. \n\n1*-18 = -18, but 1-18 = -17, so that won't work\n2*-9 = -18, and 1-9 = -7 -- THIS ONE WORKS!! yay!\n\nSo now you can rewrite your problem as\n\n3x^2 - 9x + 2x -6 = 0 and group the first 2 and last 2 terms\n\n(3x^2-9x) + (2x-6) = 0 -- you can factor a 3x out of the first group, and a 2 out of the second\n\n3x(x-3) + 2(x-3)=0 -- and now you can factor an (x-3) out\n\n(x-3)(3x+2)=0\n\n~~~~~~~~~~~~~~~~~~~\nOops, you're right. I was typing too fast.  Oh well, I still got the right final answer :)  Thanks
224079
try this site... it shows step by step how to go about sloving these equations, good luck:
891080
1. Xc = 1/2πfC\nSo 2πfC = 1/Xc\nSo C = 1/2πfXc\n\n2. ib = ibq + √2is\nSo √ = ib - ibq\nSo 2is = (ib - ibq)²\nie is = ½(ib - ibq)²\n\n3. Q = XL/R\nSo XL = QR\n\n4. Lt = L1 + L2 - 2Lm\nSo 2Lm = L1 + L2 - Lt\nie Lm = ½(L1 + L2 - Lt)\n\n5. R1=i2 R2 / i2\n??????? So R1 = R2\n\n6. Es = E1 + E2 + E3\nSo E2 = Es - E1 - E3\n\n7. R1 / Rx = R2 / R3\nSo Rx / R1 = R3 / R2\nSo Rx = R1 * R3/R2\n\n8. y = Mx + b\nSo Mx = y - b\nSo M = (y - b)/x\n\n9. i2=i1 R1 / R2\nSo i1 R1 = i2R2\nie i1 = i2 R2 / R1\n\nI would be extremely concerned about your progress if you find this simple algebra a serious chore!!!!!! (Especially if you are learning the electrical trade)