Can this be factored further?  I know I'm missing something.?
464534
Actually, -18 *does* have something to do with it. Try to find two numbers which multiply to give -18 and which add to give -5. In this case, it doesn't work, so the polynomial doesn't factor.\n\nConsider 8x^2-10x-3. Multiply the 8 and -3 to get\n-24. Now find to numbers which multiply to give -24 and which add to give -10. In this case -12 and 2 work. Now do this:\n\n8x^2-10x-3=\n8x^2-12x+2x-3= (using the -12+2=-10)\n4x(2x-3)+(2x-3)= (factoring 1st two and last two terms)\n(4x+1)(2x-3).\n\nThis gives a better way to factor that simple guessing. It will always work if the quadratic can be factored.
945958
(2n-1) * (2n-2) * (2n-3) * (2n-4)... * 3 * 2 * 1\n\nJust like any other factorial.
559152
are u sure its not supposed to be the other way around. Cuz that would make a little more sense.\n\nBut if not, then it would probably be:\n1 + ((4s squared - 11) / (s to the 4 - 3s squared -11))\n1 the quotient. the rest the remainder.
845673
First, you got the divisor & dividend backwards........\n\nFor addition the answer is called the sum.  Not sure about 3, 2\n\nSubtraction:  minuend - subtrahend = difference
865479
First of all is the radical in the numerator over just the 10 or over the entire 10y^3?\n\nThe only this you can factor out and cancel is a radical 2 on the top and bottom.  Other than that, you can't really simplify this problem.  Are you sure you wrote it down correctly?
615931
Try looking it up in your math book if you need more clarity
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)
530853
So where's that starting point?  Either we start (yesterday) with Mary having $785.63 we subtract $57. and $125. for 603.63.  Then we add $57.25 for a total of 660.88 (B) \nOR\nwe start we $785.63 (today) anb subtract $57. and $125 for $603.63 (C).  The $57.25 doesn't matter cause Mary did that yesterday....
64205
I'm not sure I understand the question.  Add some detail?
420710
I know it has already been answered but I"ll take a chance with something different. Ok 3 to the third power plus two would be 3 to the fifth power which is 243.\n :)
603820
Well i cannot write the whole formula here but check the link below
916165
Sorry, there isn't one.  If you meant "3x + 9" instead of 3x, then x + 3 is a common factor.
274876
by inspection I can understand another point will be...\n\n(0, 5) because if you add 1 to X you must add 2 to Y
485469
product of 8 and -3  = -24\nfactors for -24 are 1)-12 ,2\n                    2) 12, -2\n                    3) 4 , -6\n                    4) -4 , 6\n                    5) 8, -3\n                    6) 3, -8\n\nOf the above only the factors 1), that is -12, 2 when added give -10.\n\nTherefore we would take those factors. And split the middle term to get the following:\n\n               -3x^2 -12x + 2x + 8\n\nNow factorize to get:   -3x( x + 4 ) + 2 ( x + 4 )\n                      = (-3x + 2)( x + 4)
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
611167
You can't factor that equation.
87985
The expression was not set equal to anything.\n\nIf you're doing simple factoring in algebra class, the \nsimplest factors would be (b+4) (b+3).\n\nIf the expression was set to equal 0, solve for b, then your answer set would include -4 and -3.\n\nHope this helps!