Can someone please explain IN DETAIL how to do this problem?\n\nx^2 - 5mx + 6m^2 = 0\nWhen factored, the problem is:\n(x - 3m)(x - 2m) = 0\nlet m = 3
201108
You've done 90% of work.\n\n(x - 3m)(x - 2m) = 0\nIt can take place only if either x - 3m = 0 or x - 2m = 0.\nFirst equation means that x = 3m.\nSecond means that x = 2m.\n\nSo, there are two roots of equation: 3m and 2m.\nFor m = 3 we have roots: x = 9 and x = 6.
289766
I think you set the equation to zero then you have to factor it.  Go to the link below if you wanna know how to do it. OR look your your math book.
417398
x³y - 3x²y - 10xy --- Take x and y out of each term...\nxy(x² -3x - 10) --- Factor the x² - 3x - 10\nxy(x - 5)(x + 2)\n\n*Note: To show exponents, such as x², use the ^ key (such as x^2). This expression looks like you basiclly wrote:\n3xy - 6xy - 10xy
524529
On problem 5\n\n4X^2 + 5x - 6\n\nLook at your factors for a & c.\n4 has factors  1*4 and 2*2\n6 has factors  1*6 and 2*3.\nNow multiply the choices and find differences until a diff is 5.\n1,4 times 1,6 gives us 1-24 = -23\n1,4 times 6,1 gives us 6-4 = 2\n1,4 times 2,3 gives us 2-12 = -10  \n1,4 times 3,2 gives us 3-8 = -5..... (yea!!)\n\nUse those values in your answer 1,4 and 3,2 such that the 3 is multiplied by 1x and 2 is by the 4x.\n \n(4x-3)(x+2)\n-----------------------------------------------\nOn problem 7, \n9X^2 +12X + 4\nNote that we have positive signs on b & c.\n\nFactors of 9 are 1,9 and 3,3\nfactor of 4 are 1,4 and 2,2\nNow multiply the factors and find when the SUM is 12.\n1,9 times 1,4 leads to 1 + 36 = 37\n1,9 times 4,1 leads to 4 + 9 = 13\n3,3 times 1,4 leads to 3 + 12 = 15\n1,9 times 2,2 leads to 2 + 18 = 20\n3,3 times 2,2 leads to 6 + 6 = 12 (yea!!)\n\n(3X + 2)(3X + 2)\n\n-------------------------\nOn problem 9, \n12X^2 - 24X - 9\n\nwe have a negative c again so we will find differences between factor products until finding 24.\n\n12 factors into 1,12.....2,6.... and 3,4\n9 factors into 1,9 and 3,3\n\n1,12 times 1,9 leads to 1 - 108 = -107\n1,12 times 9,1 leads to 9 - 12 = -3\n1,12 times 3,3 leads to 3 - 36 = -33\n2,6  times 1,9 leads to 2 - 54 = -52\n2,6  times 9,1 leads to 18 - 6 = 12\n2,6  times 3,3 leads to 6 - 18 = -12\n3,4  times 1,9 leads to 3 - 36 = -33\n3,4  times 9,1 leads to 27 - 4 = 23\n3,4  times 3,3 leads to 9 - 12 = -3\n\nrats.   no solution
526072
x^2-5x-6\n\nfactored is .....\n\n(x-6)(x+1)\n\nif you want to solve this quadratic equation, set it to equal zero.\nThen factor like I did above.  So the answers will be 6 and -1
478942
College alg help?\nsolve the equation by factoring\n\n3x^2+4x-15=0\n\n3x^2+9x-5x-15=0\n(3x-5)(x+3)=0\n3x=5 or x=5/3\nx=3=0 or x=-3\n\n2.4\n__\n2+3i\nmultiplying by 2-3i\n8-12i/(4+9)\n=(8/13)+(-4/3)i\n\n3.square root of -25 in a+bi form\n+/-5i\n0+5i or 0-5i\n\n4.x^2-4x-3=0\nx=[4+/-rt(16+12)]/2\n=[4+/-2rt7)]/2\n=2+/-rt7\n\n5.22x+11=(x+6)^2\n22x+11=x^2+12x+36\n=>x^2-10x+25=0\n=(x-5)^2=0\nx=coincidental roots5,5
430968
Can this be factored?\n\n(-2x + 3)(x + 2)\n\nThis is a vertical parabola with points at (3/2, 0) and (-2, 0). The (vertical) axis of symmetry is halfway between, at\n\n(3/2 - 2)/2, or -1/4.\n\nWe can check by taking the first derivative, \n\n-4x - 1 = 0\n\nand you see that the vertex has x = -1/4.\n\nThe axis of symmetry is x = -1/4.
430061
m= -2\ny=mx+b\n5= -2(2)+b\n5= -4+b\nb=9\ny= -2x+9
658238
You know the first term is x^2 so you can write:\n(x     )(x     )\nYou know that the second term is -4x:\nIf factors are of the form (x+A)(x+B) then  Ax+Bx = -4x\nYou also know that the last term is -5, so A*B=-5\n\nJust find the values of A and B that satisfy these equations:\nAx+Bx = -4x\nA*B = -5\n\nA = -5 and B = 1\n\nNOW that we have the factors\n(x-5)(x+1)=0\nTo get 0 either section of the left hand side can be 0, so solve both cases:\nCASE 1:\nx-5 =0\nx = 5\nCASE 2:\nx+1=0\nx=-1\n\nANSWER x=5 or x=-1
224961
x=\n3x-3+2x=5x-6-x\n3x+2x-5x+x=-6+3\n1x=-3\nx=-3&#92;1\nx=-3
508471
First, factor out 3x from the equation, so 0= 3x(x^2-2x-3).\n\nThen, factor the polynomial in the parenthesis, so 0= 3x(x-3)(x+1)\n\nAfter that, use the zero product property to set each part of the the equation to zero, so 0=3x, 0=x-3, and 0=x+1.\n\nSolve these three to get x=0, x=3, and x=-1, respectively.
105050
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
406320
x^2-6x+9=3+9\n(x-3)^2=+/-rt12\nx-3=+/-2rt3\nx=3+/-2rt3\n\n2.2(x^2+5x)=-11\n2(x^2+5x+25/4-25/4)=-11\n2(x+5/2)^2-25/2=-22/2\n2(x+5/2)^2=3/2\n(x+5/2)^2=3/4\nx+5/2=+/-(rt3)/2\nx=-5/2+/-(rt3)/2\n=1/2[-5+/-rt3]
527012
Basically what it is asking you for is to factor it out. To do so you need to find two numbers that multiply to -14, and add to -5. These two numbers would be -7 and 2. \n\nThat means you would factor it as such:\n(x-7)(x+2)=0\n\nThere are two solutions:\nx=7, and x=-2
425728
2.5 m= - 25.. are you trying to solve for M.... \nand -3.25t=-3.11 i am assuming you are trying to solve for T?\n\nfor the first one... 2.5m/2.5=-25/2.5 so m = 10\n\n2nd one .. -3.25t/-3.25= -3.11/-3.25\n\ni don't know if this is what u looking for
417517
If m equals to 7\n3 (m-5)+1 = 2(m+1)-9\n3m-15+1 = 2m+2-9\n3m-14 = 2m-7 (add 7 to both side)\n3m-7 = 2m (subract 3m from both side)\n-7 = -1m\n-7 = m
635819
1. Add 3 to both sides to set equal to zero, then factor to get something that adds to three and multiplies to 28---7 and -4\n(x-4)(x+7) = 0\nx=4, 7\n\n2. Add 4 to both sides to set equal to zero, then factor to get something that adds to 2 and multiplies to -15---5 and -3\n(x+5)(x-3) = 0\nx = -5, 3