Can x2 + 3x be factored?
408554
yes. take out an x \nyou answer should be x(x+3)
609345
The demoninator of the second term (x^2 + 7x + 6) can be factored to (x+6)(x+1), so multiply both sides by that value and watch the terms cancel out.  You get \n\n5(x+1) + 2 = 3(x+6)\n5x + 7 = 3x + 18\n2x = 11\nx = 5.5
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
159156
factoring, then:\n\n(2x-3)(x+1)=0\n\nyou can assume that\n\n2x-3 = 0, and\n\nx+1 = 0\n\nSolve for first x,\n\n2x=3 , x=3/2\n\nsecond x,\nx= -1\n\nThe sum would be 3/2 - 1, or 3/2 - 2/2 =1/2
1070097
(x+3) (5x-3) (7x-1)\n\nYou can do this manually by figuring out the factors of 35 and 9.\n\nFor 35:  {1, 5, 7, 35, -1, -5, -7, -35)\nFor 9: {1, 3, 9, -1, -3, -9)\n\nI started assuming that (x + n) would be a factor.  I tried (x + 1), (x - 1), then (x + 3).\n\nIf you do long division, you get:\n(x + 3) ( 35x^2 - 26x^2 + 3)\n\nThen you just repeat the process for ( 35x^2 - 26x^2 + 3)... you need factors of 35 and factors of 3 that will add up to -26.  \n-3 x 7 = -21 and -1 x 5 = -5, so (5x - 3)(7x - 1) works.\n\nThat gives the final answer of: \n(x+3) (5x-3) (7x-1)
916165
Sorry, there isn't one.  If you meant "3x + 9" instead of 3x, then x + 3 is a common factor.
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)
914755
your equation must be like x^2-3x-6 and your trying to factor this\n\n2 numbers say a and b must multiply to make -6 and add to make -3 so a*b=-3 and a+b=-6\n but i can't think on any that fits this criteria so you must use the quadratic equation\n\nx= [-b +-squareroot(b^2-4ac)]/2a\nwhere a=1 b=-3 and c=-6\nx= 3+-squareroot(9-4(1)(-6))/2\n\nx=(3+squareroot(33))/2 and x= 3-squareroot(33)/2\n\nin decimals x=4.3723 and x=-1.3723
600427
It is already factored, you can write it more concisely as (x-y)^2.\n\nThe EXPANDED form is x^2-2xy+y^2.
913289
You end up with a remainder, so it's not pretty, but you should be able to do long division and get:\n\n3x^2 - 4x + (16/3) - 23/(3*(3x+2))
818959
Substitute x + 7x in for 8x:\n\nx² + x + 7x + 7\n\nNow partial factor out the common terms:\n\nx(x + 1) + 7(x + 1)\n\nnow factor out the (x + 1):\n\n(x + 1) * (x + 7)
1174486
First look at the highest powers of a and b which in this case is "squared" and the product 6ab is present, so factoring will likely involve (a+b)^2\nIn fact 3a^2+6ab+3b^2 = 3(a^2+2ab+b^2) = 3(a+b)^2\nSo your problem reduces to:\n\n3(a+b)^2 +7(a+b) - 6\n\nIf we choose to factor -6 as -2 x 3 then the it is easy to factor, and the solution is:\n\n(3(a+b) -2)((a+b)+3)
838099
Yes, you can factor it as follows:\nx² - 22x + 7x \n= (x - 4)(x - 18)
119973
(x+3)(x^2-x+4)/(x+3) common factor\n\nmultiply by x+3\n\nx^2-x+4
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~ ♥ ~
836052
You can start by factorizing 3x^4 from everything\nm(x) = 3x^4 (2x^3 + 5x^2 - 3x - 3)\n\nI think there's something wrong with your problem, none of the possible roots given by the Ruffini's rule are working. Perhaps, if instead of 6x^7 + 15x^6 - 9x^5 - 9x^4 it was 15x^7 + 15x^6 - 9x^5 - 9x^4, then it could be factorized very nicely in\nm(x) = 3x^4 (x+1)(5x^2-3) \nor if instead you wrote 6x^7 + 6x^6 - 9x^5 - 9x^4 then it'd factorized to\nm(x) = 3x^4 (x+1)(2x^2-3)
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