Hello,  I'm in algebra 2 and i wanted to know how to completely factor this:   (a+b)^6 - (a-b)^6 .  Thanks!!
822596
Note that this is a difference of squares, so all you have to do is factor is as a difference of squares.\n\nReminder:   x^2 - y^2 = (x-y)(x+y)\n\nIn this case, (a+b)^6 is the same as ((a+b)^3)^2\n\nSo (a+b)^6 - (a-b)^6 = \n\n[ (a+b)^3 - (a-b)^3 ] [ (a+b)^3 + (a+b)^3 ]\n\nNow we have sums and differences of cubes; the proper way to factor sums and differences of cubes are:  \n\n(x^3 + y^3) = (x + y) (x^2 - xy + y^2)\n\nVerbally, the tactic would be:  "square the first" "negative product" "square the last".\n\n[ (a+b)^3 - (a-b)^3 ] [ (a+b)^3 + (a+b)^3 ]    becomes\n[ (a+b) - (a-b) ] [ (a+b)^2 + (a+b)(a-b) + (a-b)^2 ] [ (a+b) + (a-b) ] [ (a+b)^2 - (a+b)(a-b) + (a-b)^2 ]\n\nsimplifying, we get\n\n(2b) [(a+b)^2 + (a+b)(a-b) + (a-b)^2] (2a) [ (a+b)^2 - (a+b)(a-b) + (a-b)^2 ]\n\n(4ab) [a^2 + 2ab + b^2 + a^2 - b^2 + a^2 - 2ab + b^2] [ (a+b)^2 - (a+b)(a-b) + (a-b)^2 ]\n\n(4ab) [3a^2 + b^2] [a^2 + 2ab + b^2 - (a^2 - b^2) + a^2 - 2ab + b^2]\n\n(4ab) [3a^2 + b^2] [3b^2 + a^2]\n\nThis can't be factored any further.
988729
(a + b)^2 = a^2 + 2ab + b^2\n\nso\n\n(a + 1)^2 = a^2 + 2a + 1\n(a - 1)^2 = a^2 - 2a + 1\n\n\n\nFermat:\n\nIf a = 2^(n-1) then\n(a + 1)^2 = (2^(n-1) + 1)^2\n = (2^(n-1))^2 + 2(2^(n-1)) + 1\n =( 2^(n-1)^2 ) + (2^n + 1)\n\n\n\nMersenne:\n\n(2^n - 1) + (2^(n-1) - 1)^2 =\n = 2^n - 1 + (2^(n-1))^2 - 2(2^(n-1)) + 1)\n = 2^n - 1 + (2^(n-1))^2 - 2^n + 1\n = (2^(n-1))^2\n = ( (2^(n-1) - 1) + 1 )^2
234829
With factorisation you turn the equation x^2 + px + q = 0 into two smaller equations in brackets like this:\n\n(x+a)(x+b) = 0\n\nSo that a+b = p\nand a*b = q\n\nEG. Your first example you have p = 4 and q = 3\n\nSo you've got a+b = 4 so a could be 2 and b 2 or a 1 and b 3 etc\n\nand a*b = 3 so a could be 1 and b 3... Yes it matches.\n\n(x+3)(x+1) = 0\n\nThe solution is just x = -a and x = -b\n\n-> x = -1 or x = -3
693658
x2 + 5x + 6 = 0\n\nto factor this, what times what equals 6, and when added together, equals 5?\n\nwhen worked out, you come to the conclusion that 3 x 2 = 6, and 3 + 2 = 5.
703982
1.\n3x + 6/((x + 1)(x + 2)) Factor 3 out of the top.\n[3(x + 2)]/[(x + 1)(x + 2)] x+2's cancel\n3/(x + 1) <-- This is simplified.\n\nI don't have the energy to do the other ones right now.  :/
529729
(x2)-3 (x-2) over (x2) -4?\nIs this what you mean?\n(x^2)^(-3)*x^(-2)/(x^2)^(-4)\n= x^(-6)*x^(-2)/x^(-8)\n= x^(-6-2-(-8))\n= x^(-8+8)\n= x^0 \n= 1
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=9x^2+6x+1\n\nsince (a+b)^2=a^2+2ab+b^2\n\nhttp://www.algebrahelp.com/lessons/
930643
factor:\n\n(2x^2+x-6)/(2x+4)=\n\n(2x-3)(x+2) / 2(x+2)\n\nthe (x+2)s cancel out leaving\n\n(2x-3)/2
503348
No, the previous answer is wrong.\n\nGroup it like this:\n\n2p + pq -6q -3q^2\n\nthen factor like this:\np(2+q) -3q(2+q)\n\nfactor out the 2+q to get:\n(2+q)(p-3q)\n\nFOIL it out and you will see it is correct.
579949
Factor question:\n\nLet y^1/3 be x.\n2x²+x-6 = (2x-3)(x+2)\nTherefore the answer is (2y^1/3+3)(y^1/3+2)\n\nSolve #1:\n\nLet x^2 be "x".\n3x²-5x-2 = 0\nUsing the quadratic formula, we get (5(+-)sqrt(25+24))/-5, or either -2.4 or 2/5. So the answer is either 5.76 or 0.16.\n\nSolve #2:\n\nx²+6x+1 = 0\nUsing the quadratic formula, we get (-12(+-)sqrt(36-4))/12, or (either -1 - sqrt(32)/12 or -1 + sqrt(32)/12).\n\nHope this helps!
820930
First, factorise the y²+y-6 using the quadratic equation resolution:\ny = (-b ± sqr(b² - 4ac)) / 2a\ny = (-1 ± sqr(1 - 4.1.-6)) / 2.1\ny1 = -3, y2 = 2, so the factors of the second denominator are\n(y + 3) and (y - 2).\nWow!  Look!  the first denominator is (y - 2)!\nSo we just have to multiply the first fraction by (y+3)!\n[1 * (y+3) / (y-2) * (y+3)] + [3 / (y-2) * (y+3)]\nWe have a common denominator, so we just have to add the numerators...  That you can do...
833702
Do you mean (2x^2 + 3x + 1/ 2x^2 + x - 3) for your first fraction?  I'm going to assume you do.\n\nFirst, factor each numerator and denominator:\n\n2x^2 + 3x + 1 = (2x+1)(x+1)\n\n2x^2 + x - 3 = (2x-3)(x+1)\n\n4x - 4 = 2(2x-2)\n\nNow let's look out our multiplied fraction, rewritten using our factors we just found:\n\n[ (2x+1)(x+1)2(2x-2) ] / [ (2x-3)(x+1)(2x+1) ]\n\nNotice we can cancel out the (2x+1)'s and the (x+1)'s, to get:\n\n2(2x-2)/2x-3\n\nANSWER:   2(2x-2)/2x-3
636732
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 :)
470305
Factoring is the "opposite" of FOILing.\n\nIf you had x^2 (which is x squared)\nx^2 + 5x + 6\n\nYou ask yourself:  What multiplies to 6?  (Write all the choices including negatives)\nThen, out of those choices, which one (and only one will work) adds to 5.  \n\nSo... 2 x 3 = 6  and -2 x -3 = 6\n2 + 3 = 5\n-2 + -3 = -5 \n\nSo, 2 and 3 add to 5 and mult. to 6.\n\nThen you write it in parenthesis keeping the signs.\n\nYou get (x + 2)(x + 3)\n\nHope that helps.  Good luck.
972642
(A^2+2ab+b^2)/(a+b) + (a^2-b^2)/(a-b) + (a-b)/(a+b)/(a^2-b^2)/a+b)\n(a+b)^2/(a+b)+(a+b)(a-b)/(a-b)+ (a-b)/(a+b)*(a+b)/(a^2-b^2)\na+b+a+b+(a-b)/((a+b)(a-b))\n2a+2b+1/(a-b)\n\nif I got all the ( ) in the right places.
528285
From\n6a^2 + 12ab - 18b2\nFirst, we factor out the common factor 6 to obtain:\n\n6(a^2 + 2ab - 3b^2)\n\nsince, the third term is - (negative) our factors will have different signs;\n6(___ - ___  )(___ + ___).\n\nThe first term's coefficient is 1, whose only factor is 1, so our factors will look like this:\n6(a - ___)(a  + ___)\n\nThe third term is -3b^2, and its coefficient is -3, whose factors are (1, -3), (-1, 3). Of these we use the factors such that when added will give us 2 (which is the coefficient of the second term 2ab. So we choose (-1, 3)\n6(a - b)(a + 3b)
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~ ♥ ~