factoring trinomials!!!!!!?
970180
Use the slide method!\n2x^2 + 5x + 12\nMultiply the coefficient of x^2 with the the last term (the constant C), so it becomes:\nx^2 + 5x + 24\n(x + ) (x + )\nSince they're both positive, there are no factors of 24 you can plug into that would add up to 5.  Thus, you can't factor it.\n\n-4x^2 -9x - 10\nAgain, multiply the coefficient of x^2 with the constant.\nx^2 - 9x + 40 (it becomes positive because -10 * -4 = -40)\nThe plus sign after the 9x tells you that the signs are both going to be the same, and the subtraction sign after the x^2 tells you that they're both going to be subtraction signs.\n(x- )(x- )\nThen, think of what factors of 40 add up to 9.  None, so this can't be factored either.\n\n4a^2 - a - 5\nOnce again, multiply the coefficient of a^2 with the constant.\na^2 - a - 20\nYou can see from the - sign after the a that the signs will be different.\n(a+ ) (a- )\nThen think of factors of 20 that will equal -1.  \n(a+4) (a-5)\nFor it  to be -1, the bigger number has to be negative, so it will be a-5.\nBut then we have the problem of the -4 (coefficient of a^2) we got rid of earlier.  So, we have to divide the numbers by -4:\n(a + (4/-4))(a - (5/-4))\n(a + (4/4)) = (a - 1)\n(a - 5/-44)) doesn't divide cleanly, so the denominator goes in front of a. Thus:\n(a - 5/-4)) = (4a + 5)\n\nYes, no solution is definitely possible.  And you can use the discriminant to check:\nb^2 - 4 * a * c\na is the coefficient of x^2, b is the coefficient of x, and c is the constant.  If you plug those into the equation and you come up with a negative number, then there is no solution.\n\nYou should be able to do the last one.\nBasic rules of the slide method:\n-Multiply the coefficient of x^2 with the constant\n-Factor\n-Divide constants by coefficient: if it doesn't divide cleanly, simplify if possible and put whatever is left of denominator in front of the x.  For example:\n(x + 4/6) = (3x + 2) (because you can simplify both 4 and 6 by dividing by 2)
503153
(x+7)(x+3)\n(x-11)(x+2)\nforget this one.  its AC factoring though\nthis ones AC too
241036
I'm assuming you're looking for information on factoring polynomials since it's a College Algebra courese.  \nThis website is very helpful, with the steps for factoring laid out, and it also provides websites you can use if you need more information.  Good luck!\n\nhttp://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut7_factor.htm
195306
I haven't done factoring for a while but I'm pretty sure this is right.\n\nSo you start with 125q²-n³q²\nIf you take out both the q² 's you have q^4 \n(q x q x q x q)\nSo you are left with q^4(125-n³)\nThe cubed root of 125 is 5\n(5 x 5 x 5)\nand the cubed root of n³ is n\n(n x n x n)\nSo your answer is q^4(5-n)\n\nI think that's right.
594239
ok so you have p^2 + p, you can factor out p from both parts of the expression, it will end up like this: p(p + 1).  Hope that works.
621145
With just ! (Factorial), +, -, * and /, it is impossible to get 150 using only 3 5s without the x^2 (Squared) operation (Which means x*x, in case you are in Fifth Grade), or using a '^2' as you say. If you say there is an alternate way, and there really is, then I give up and admit defeat. (I'm sure most others do.) If you are so sure, why don't you post the answer yourself?
1027973
the 3rd one is correct, but I get -(3+b)/b+2 and 2 for the first 2
255329
Have patience :)  You got some answers the first time you asked this.  Here it is again:\n\nIt 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 1-9 = -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\n~~~~~~~~~~~~~~~~~~~\nOops, you're right. I was typing too fast.  Oh well, I still got the right final answer :)  Thanks
36762
Factor by grouping:\nx^2 (3x - 5) - 3 (3x - 5)\nAnswer: (3x - 5)(x^2 - 3)
91753
factorize\nx^2 + x = x(x+1)\n\nfactorize\nx^2 - 1 = .... (you should know)\n\n2 = 2 (!!!)\nTake the lcm of the three denominators (each different factor (=parenthesys) once and with the highest power)\n\nIt is 2x(x+1)(...)   (fill the dots)\n\n.....\n\nnow if you can't go on, you need to open your book again and study, or pay more attention at school at your math classes, and ask immediately for further explanation to your teacher when you feel you've not understood.\n\nGood work and good study
908426
Plug in each of the four answers.  The only one that works is (2), because you get\n\n | 3 (-2/3) - 3 | = 5\n | -2 - 3 | = 5\n | -5 | = 5\n 5 = 5
835153
.more kids homework... factoring?\n1. rewrite the middle term as the sum of two terms and then factor completely 15x^2 -21x +6\n\n3(5x^2-7x+2)\n3(5x^2-5x-2x+2)\n3(5x-2)(x-1)\n\n2. rewrite the middle term as the sum of two terms and then facto by grouping. x^2 - 14x +45 \n\nx^2-9x-5x+45\n(x-9)(x-5)\n\n3. rerewrite the middle term as the sum of two terms and then facto by grouping x^2 + 19x +90\n\nx^2-10x+9x+90\n(x+10)(x+9)\n\n4. rewrite the middle term as the sum of two terms and then facto by grouping x^2 -2x -63\n\nx^2-9x+7x-63\n(x-9)(x+7)\n\n5.solve x^2 = -6x\n\nx=-6
952644
In a perfect square trinomial the middle term ("x" term) is twice the product of the two terms in the factors (ax+b)^2 = a^2x^2 +2*a*b*x+ b^2.  So if  the trinomial is cx^2 + dx + e,  a=√c, b=√e.  So a = ±6, and b = ±6.  You  have  to decide whether they are both + or -; since 2*a*b= -72, so a*b = -36 so the factors must be (6x - 6)*(6x - 6).\n\nEDIT:  Naturally to complete the factorization, you would separate the constant factor in each term: 6*(x-1)*6*(x-1) = 36*(x-1)*(x-1); I left it in at the start to illustrate the general approach to perfect-square trinomials.
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)
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
436598
Remember that a factorial is basically the product of that number, that number - 1, all the way down to 1.\n\n1. (n-1)(n-2)(n-3)(n-4)(n-5)...1\n---------------------------------\n(n-4)(n-5)...1\n\nCancel the (n-4) and everything after.\n\nANSWER : (n-1)(n-2)(n-3)\n\n2. (n+6)(n+5)(n+4)(n+3)...1\n---------------------------\n(n+3)(n+2)...1\n\nCancel the (n+3) and everything after\n\nANSWER : (n+6)(n+5)(n+4)\n\nEDIT : Lattisrectum, keep in mind that (n-4)! is (n-4)(n-5)(n-6)... all the way to 1, not (n-4)(n-3)... :)
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.