How do you factor this?
372976
yes u can\nt^2(s^4-4)is the factored form\nt^2 is the highest common factor\n\n\n75x^2 - 147y^2\n3 is the common factor\n3(25x^2-49y^2)\n3(5x^2-7y^2)(5x^2+7y^2) is the factored form of 75x^2 - 147y^2
977376
at the first 4 you use a^3-b^3=(a-b)(a^2+ab+b^2)\n and a^3+b^3=(a+b)(a^2-ab+b^2)\n\nfor the next 2 :\nthe first 2 factors and the second 2 factors have something in comon\n\nthe next ones use the above formulas plus one more \na^2-b^2=(a-b)(a+b)
287286
This is a quadratic equation hence there are two factors.\n\nFirst factorise the equation: (3x-1)(2x-3)=0\nThen each factor = 0\nHence x=1/3 (from the first factor)\nand x=3/2 (from the second factor)\n\nyou can recheck this by substituting these two values in the original equation and it shoudl add up to zero.\n\nThe factors are basically points on the curve that cut the x axis (where y=0)
798535
x^3 - x^2 + 5x + 5\n\n(x^3 - x^2) + (5x - 5)\n\nx^2(x - 1) + 5(x - 1)\n\nsince both x^2 and 5 have the same factor, you can factor it to\n\n(x^2 + 5)(x - 1)\n\nif you want, you can either leave it that way, or\n\n(x + sqrt(5i))(x - sqrt(5i))(x - 1)\n\nwhat you can probably just leave it as (x^2 + 5)(x - 1)\n\nyou can also try www.quickmath.com, click Factor under Algebra, then type in x^3 - x^2 + 5x - 5, then click enter. \n\nAnymore Questions Ask ME
1119411
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
684474
You have to factor each part first, then cross off (cancel) identical factors from the numerator and denominator of either expression.  If the factors are exact opposites, like x-8 and 8-x, cancel and put a -1 in the numerator.\n\n\nIn your example only the first expression is factorable:\n(x-8)(x+3)/(2x)(3+x) times 5x^2/(8-x)\n\nThe (x-8) cancels the (8-x) to become -1, then (x+3) cancels the (3+x), and the x from the (2x) cancels one of the x's from (5x^2)\n \nAnswer: -5x over 2
791576
I'm going to assume the placement of parenthesis.\n(x^2 – 3x + 2)/(8x - 8) ÷ (x^2 – 4)/(5x +10)\n\nStep 1: Flip the second fraction, so that you have multiplication instead of division.\n(x^2 – 3x + 2)/(8x - 8) * (5x +10)/(x^2 – 4)\n\nStep 2: Factor.\nx^2 – 3x + 2 = (x - 2)(x - 1)\n8x - 8 = 8(x-1)\n5x +10 = 5(x + 2)\nx^2 – 4 = for a^2 - b^2, the factors are (a + b)(a - b). So: (x + 2)(x -2)\n\n(x - 2)(x - 1) / 8(x-1) * 5(x + 2) / (x + 2)(x -2)\n\nStep 3: Cancel out, multiply\n(x - 2) / 8 * 5 / (x -2)\n5 / 8 (solution!)
1002101
factor 9x^2y^2-9y^2\n9y^2(x^2-1)\n9y^2(x+1)(x-1)
1053403
Write it as:\n2x^2-4x+3x-6\n=2x[x-2]+3[x-2]\n=[2x+3][x-2]\nSimple middle-term factorisation!!!
508408
It's called factoring by grouping. Take a factor out of two of the four terms (and the same for the other two) so they share a common factor.\n(2x^3 - x^2) - (18x - 9)\nx^2(2x - 1) - 9(2x - 1)\n(x^2 - 9)*(2x - 1)\n(x + 3)*(x - 3)*(2x - 1)\n(ungkog WAS correct)
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)
727014
start with the O... they are always -2\nsince there are 2, and the compound must balance...\nPb must be +4\n\nso you get +4 -2
796585
1.factor\n(2x+1)(x-4)=0\n2.set it to 0\n2x+1=0\nor\nx-4=0\n3.solve\n{x=-1/2 or x=4} <=answer
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!
283601
1) First we move all the factors with x to one side, and the ones without x to the other.  Result will be:\n\n4x/3 - 2x + x = 3/2 + 1/5\n\n2)  Then, we factor out an x from the left side and get:\n\nx(4/3 - 2 + 1) = 3/2 + 1/5\n\n3)  Calculate the actual numbers on both sides\n\nx(4/3 - 6/3 + 3/3) = 15/10 + 2/10\nx(1/3) = 17/10\n\n4)  Divide both sides by 1/3 and get a final result of...\n\nx = (17/10)/(1/3)\nx = (10/17)*(3/1)\nx = 30/17\n\n5)  Comform that I PWN n00bs.  And... I'm pretty sure I pwn n00bs.  Therefor, we are done!  ^-^
614345
(x+10)(x-10)\n(x-4)(x²+4x+16)\n(x+w)(x²-wx+w²)\n\nUnfortunately you just have to memorize the formulas.  There is no other way to figure out the factoring of the cubic expressions.
842037
We need to factor the equation down. This involves finding the roots of the equation. This will look like:\n\n(ax+b)(cx+d) = 12x^2-31x+20 = 0\n\nBecause the final term is positive and the middle term is negative, my first hunch is to have b = -4 and d = -5. By doing so, the equation looks like:\n\n(ax-4)(cx-5) = 12x^2-31x+20 = 0\n\nThe factors needed for 12x^2 could be 2 and 6 or 3 and 4. We should test each combination to see that they work out. The factoring ends up looking like this:\n\n(3x-4)(4x-5) = 0\n\nNow all that remains is to solve for x in each set of parentheses. If the first term is zero, then\n\n3x-4 = 0\nx = 4/3\n\nIf the second is zero, then\n\n4x-5 = 0\nx = 5/4\n\nThus, x = 5/4, 4/3