6x^2-14x-12=0 I need help with factoring this, and I need to know how do you chose which factors to use, and what happens to the 14x in this problem.  How do you eliminate it?
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It 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 2-9 = -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\nEdit: correcting typos :)
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(1) any of the following may be the root of the expression,+1,-1,+2,-2,+3,-3,+6 or -6.\nNow putting the value of x=-1,see ifthe value of the expression becomes zero or not?If it becomes zero,then x=1 is one factor of the expression,if it does not become zero try with =1,then -2.+2 and so on.\nLuckily for us,with x=-1,the value of the expression becomes zero\nTherfore,x+1 is one factor\nx^4++x^3-7x^2-13x-6\n=x^3(x+1)-7x(x+1)-6(x+1)\n=(x+1)(x^3-7x-6)\n=(x+1){x^2(x+1)-x(x+1)-6(x+1)}\n=(x+1)(x+1)(x^2-x-6)\n=(x+1)^2(x-3)(x+2)\n(2)Will you please check up the second one once more and if there is any correction send it to me through Yahoo answers.
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The equation is a quadratic, (you know this this because of the x squared) To solve a quadratic you need to factoruse ut, so make two brackets with and x in each, (x   )(x   ).\nNow, you need to look for factor pairs of 36(the constant at the end) that when you add them together or take them away, they equal 12 (the number in front of the x term)  Our options are, 1x36, 2x18, 3x12, 4x9 and 6x6.  Only 6x6 adds to give us 12.  You can now add two sixes to your brackets, so (x   6)(x   6).  Now you need to look at the signs, which signs when multiploed out will give us +36?\n\nIf you are not confident with algebra, try trial and error.  in this case it'll be ++, \n\nSo fully factorised this gives us (x+6)(x+6)\n\nTo solve this we now need to make it equal to 0\n\n(x+6)(x+6)=0\n\nNow put each bracket equal to zero  ( as both brackets are the same we need only look at one of them)..(x+6)=0, so with simple manipulation we can say that x=-6\n\nThe second equation is related to a circle and tricky to explain by message, try looking in a revision book around circle equations and properties
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8x^2y^3+5y^2 when "FACTORED COMPLETELY"=\n\ny^2 (8yx^2+5). \n\nThis should be right. My dad is a complete genius and he did this in like 2 seconds. I double checked it on my calculator too by plugging in numbers, so I can assure you he's right.\n\nHope this helps! Have a nice day :)
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OK First you take out the common factor which is y. After you have done that.you have (y^2-12y+36). After this, it becomes simple factoring. The answer is\ny(y-6)(y-6)\nor also known as y(y-6)^2
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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.
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What??\n\n3x2 - x +6 = 0\n6 -x +6 = 0\n12 - x = 0\nwhen x is 12 the equation is zero.\nwhen x is less than 12, the equation is less than zero.\n\n\n\n(-1)2 - 4(3)(6) = x\n-2 - 72 = x\n-74 = x
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1. Move the constant to the right side:\nx^2 - 6x = 4\n\n2. Take half of the x coefficient (in this case, -6), square it, and add it to both sides:\nx^2 -6x + 9 = 13\n\n3. Factor the left side to (x + a)^2, where a is the number you just squared:\n(x - 3)^2 = 13\n\n4. Take the square root of both sides, remembering that this will give you both a positive and a negative root:\nx - 3 = +/- sqrt 13\n\n5. Move the constant on the left to the right:\nx = 3 +/- sqrt 13
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(x-7)(x-7)\nuse the FOIL method (first outer inner last)\nx times x= x squared\nx times -7 =-7x\n-7 times x =-7x\n-7 times -7 = 49\nxsqured-14x+49\nthats factoring, but u cant really solve cuz u didnt put an equation...\n\nunderstand?
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4x-2x-6=8x-12\nadding 6\n2x=8x-6\nadding -8x\n-6x=-6\ndividing by -6\nx=1
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-6-9x=-9x-4\n-6=-4.....\nno solution?? whats so hard to that?
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2(x^2+3x-40) <<<<<The answer\n\n\nfor the guy below me...that wouldn't be factoring.
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[(15x^5)-(10x^2)+(20x)]/[5x^5]\n   find a common factor\n(5x)[(3x^4)-(2x)+(4)] / (5x)(x^4)\n   division by self equals 1/1, disregard\n[(3x^4)-(2x)+(4)] / (x^4)\n   invert and multiply (1/x^y = x^-y)\n(x^-4)[(3x^4)-(2x)+(4)]\n   Multiply through, simplify [x^y * x^z = x^(y+z)]\n3x^0 - 2x^(-2) + 4x^(-4)\n   For non-zero x, x^0 = 1\n3 - 2x^(-2) + 4x^(-4)\n\nThree minus Two x to the negative second plus Four x to the negative fourth
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first distribuit\n24x-6x=12x-36\nthen\n18x=12x-36\n-12x\n6x=-36\ndivide\nx=-6\n\nadd x\n-10>x+36\nsubtrack 36\n-46>x\nso\nx<-46
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I'm guessing that second term was to be -11x.\n\nSadly, this is trial and error.  as the last term (+24) you know the terms are both like signed (both x+something or X-something).  The second term being negative (-11x), we know that they are both x-something.  so we are looking for two factors:  (x-a) and (x-b), where:\n\n1.  a+b=11, and\n2.  axb=24\n\nOur factors of 24 are:\n1x24\n2x12\n3x8\n4x6\n\nas 3x8 fits, we try it:\n\n(x-3) (x-8)\n\nthis equals:\n\nx^2-11x+24\n\nso we are right with these two factors\n\nHope this helps....I do these things instinctually, so I have trouble explaining them.
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Actually, -18 *does* have something to do with it. Try to find two numbers which multiply to give -18 and which add to give -5. In this case, it doesn't work, so the polynomial doesn't factor.\n\nConsider 8x^2-10x-3. Multiply the 8 and -3 to get\n-24. Now find to numbers which multiply to give -24 and which add to give -10. In this case -12 and 2 work. Now do this:\n\n8x^2-10x-3=\n8x^2-12x+2x-3= (using the -12+2=-10)\n4x(2x-3)+(2x-3)= (factoring 1st two and last two terms)\n(4x+1)(2x-3).\n\nThis gives a better way to factor that simple guessing. It will always work if the quadratic can be factored.
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Factor each one.\n(x^2-3x-18) / (x^2-4x-32) divided by (x^2-2x-24) / (x^2-x-20)\n= (x^2-3x-18) / (x^2-4x-32) times (x^2-x-20)/(x^2-2x-24)\n= [(x-6)(x+3)]/[(x-8)(x+4)] times [(x-5)(x+4)/(x-6)(x+4)]\n= [(x-6)(x+3)(x-5)(x+4)] / [(x-8)(x+4)(x-6)(x+4)]\nYou can now cancel anything on top with anything on bottom\nx-6 and x+4 cancels\n= [(x+3)(x-5)]/[(x-8)(x+4)]