can someone factor 1/4b^2+4bc+bc^2?
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?
417593
no because 8n^2 - 72p^2 can still be factored\n8 is the common factor \n8(n^2-9p^2)\n8(n-3p)(n+3p)
1144353
4a+4b+4c=2ab+2bc+2ac \n4(a+b+c)=2(ab+bc+ac)\na=2\nb=2\nc=2\n\nor \na=1\nb=2\nc=4
394036
Try to solve the eqs with quad rule:\na*x^2 + b*x + c=0\nx=(-b +- Sqrt(b^2-4*a*c))/(2*a)\n\nx^2+2x+5=0 => Get two Complex answers, \nso no it can't be factorized.\n\nx^2+4x+5=0 => Get two Complex answers, \nso no it can't be factorized.\n\nx^2+4x+3=(x+3)*(x+1)=0; \nSo, yes, this can be factorized. \nSolutions are x=-3 and x= -1\n\nx^2+2x-3=(x+3)*(x-1)=0;\nSo, yes, this can be factorized. \nSols are x= -3 and x=1
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
610613
In 6ab + 4a + 3b + 2, we group the first 2 terms and the last 2 terms:\n\n6ab + 4a + 3b + 2 =\n(6ab + 4a) + (3b + 2)\nthen factor out the common factor 2a in (6ab + 4a)\n(Note: to factor out is the same as dividing each term with the common factor)\n\nwe get:\n2a(3b + 2) + (3b + 2)\nNotice that in each term of the above we have a common factor of (3b + 2)\nBy dividing each term by (3b + 2) or factoring out (3b + 2) we have:\n(3b + 2)(2a + 1)
394265
Factor it out-- A(B+C)=AB+AC.  So, let X=A, B=1, C=Y.  You get X+(X*Y)=X(1+y)=Z
271749
Yes.\n\n1.Let A = B\n\n2.Thus, by definition, A2 = AB (A2 = A squared)\n\n3.Add A2 to each side:\n\n2A2 = A2 + AB\n\n4.Subtract 2AB from each side:\n\n2A2 - 2AB = A2 + AB - 2AB, which in turn is the same as:\n\n2A2 - 2AB = A2 - AB\n\n5.Factor (A2 - AB) out of each side:\n\n2 (A2 - AB) = 1 (A2 - AB)\n\n6.Divide each side by (A2 - AB):\n\n2 (A2 - AB)    =    1 (A2 - AB)\n             _________        _________\n   (A2 - AB)    (A2 - AB)\n\n7.The result:\n\n2 = 1\n\nPS:  Obviously there is a clear logical error embedded in this. :)
272590
1/a + 1/b = c  multiply both sides by ab\n\nb + a = abc  get the terms with b on one side\n\na = abc - b factor out a b\n\nb(ac - 1) = a  divide by ac - 1\n\nb = a/(ac - 1)
762985
(b-4)/(b-4)(b+4) simplifies to 1/(b+4)\nrestrict denominator...bcannot equal 4 or -4
910180
3ab(2ab^2)^4\n3ab(16a^4b^8)\n48a^5b^9\ntrust me!!!! i just got done doing this in class a week ago.
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)
85841
How about this? Straight multiplying gives you the following:\n\n18b^3/b^2\n\nb^3 = bxbxb\nb^2 = bxb\n\nYou have 3 bs on top and 2 on bottom. Whenever you have the same number on top and bottom, it cancels out. Since there are 2 bs on bottom, 2 of the ones on the top cancel out as well. This leaves you with 18b.\n\nDid that help at all?
6182
You can use what is called the Master Product Method to factor this and other quadratic trinomials.\n\nFor something of the form ax^2 + bx + c, the master product is a*c.  In this example, a is 2, b is -5 and c is 2, so the master product is 2*2 = 4.  Now you must find two numbers which multiply to give you the master product of 4, but add to give the middle coefficient of -5.\n\nThis requires a bit of trial and error, but you should be able to quickly discover that the numbers -1 and -4 will work since -1*-4 = 4 and -1+-4 = -5.  Now you rewrite the the entire trinomial replacing the middle term -5x with the equivalent -1x - 4x:\n\n    2x^2 - 5x + 2\n= 2x^2 - 1x - 4x + 2\n\nWe factor by grouping the first pair and last pair together and pulling out what's common to each pair:\n\n= (2x^2 - 1x) + (-4x + 2)\n= x(2x - 1) -2(2x - 1)\n\nNotice that (2x-1) appears in the factorization of both pairs.  We finish by factoring out the (2x-1) to get\n\n= (x - 2)(2x -1)
504201
Actually, none of the above are correct. To factor anything in the form of a cubed - b cubed, where a and b are real numbers, the formula is (a - b)(a^2 + ab + b^2). In this case, the a would be 1 and the b would be 4. So, the answer would be (x-4)(x^2 + 4x + 16i)\n\nand paul is incorrect also, try to multiple it out, it's wrong
1109869
multiply them together to find out 7/16 x 4/3 x 1/2 = ????
882012
What you want to do is factor out the same values from the numerator and denominators.  \n\nThe order in which you factor these out does not matter, the key is to continue to factor them out until there is nothing else to factor out.\n\nFor example you can factor out a^8 by turning it into:\n\n36 * (a^8) * (b^4) * 1/2\nover\n27 * (a^8) * a * (b^6) * 1/3\n\nsince (a^8) over (a^8) = 1 you can cancel them out leaving you with:\n\n36 * (b^4) * 1/2\nover \n27 * a * (b^6) * 1/3\n\nwe do the same factoring for b where we can factor out a (b^4):\n\n36 * (b^4) * 1/2\nover \n27 * a * (b^4) * (b^2) * 1/3\n\ncanceling the (b^4)s we get:\n\n36 * 1/2\nover\n27 * a * (b^2) * 1/3\n\nYou can then further reduce the fraction by multiplying out the 36 * 1/2 and the 27 * 1/3:  leaving you with:\n\n18\nover \n9 * a * (b^2)\n\nAs you can see now you can factor out a multiple of 9 leaving you with:\n\n2\nover\na*b^2