need help on a college algebra problem! please help me!!!!?
598723
possible rational factors are p/q where, p is factors of the constant and q is factors of the leading coefficient.\np = 15 and q = 2\nfactors of p: 1, 3, 5, 15\nfactors of q: 1,2\nPossible p/q = (plus or minus) 1, 3, 5, 15, 1/2, 3/2, 5/2, and 15/2\nKeep picking one of these values and test it until you get a remainder of 0.\n1 works\n1 2 3 -11 -9 15\n ___ 2  5  -6 -15\n__2 5 -6 -15 0\nThus,\n2x^4 + 3x^3 - 11x^2 - 9x + 15 \n= (x-1)(2x^3 + 5x^2 - 6x -15)\n2x^3 + 5x^2 - 6x -15 can be factored by grouping.\n(2x^3 + 5x^2) + (- 6x -15)\n= x^2(2x+5) -3(2x+5)\n= (x^2-3)(2x+5)\nAnswer: (x-1)(x^2-3)(2x+5)
680155
i'd love to help but i'm having trouble with it myself\n\nuse ur math textbook it will have it unless ur math book say algebra and is red ( that my alg. book and it doesn't help) \n\nSRY good luck
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.  :/
937830
T=m1*-9.8+m1*a\nT=m2*a--m2*-9.8\n\nset them equal and solve for a\n\ntrust me, i had a test on this today
29264
Yuo can't solve this.Because it is a formula and not an equation..If u assign some value equal to this formula such as 0 u have the solution.
456792
www.algebra-online.com\nwww.sosmath.com/algebra/algebra.html\nmathworld.wolfram.com/topics/Algebra.html\nmathforum.org/algpow\nhttp://education.yahoo.com/homework_help/math_help/chapter_list?id=minialg1gt\nwww.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra\nmath.about.com/od/collegealgebra\n\nGOOD LUCK!! :) Happy studying!
390719
so you have (1/2)a+a-(6/3)=4+(1/3)a\n\nadding like terms: (3/2)a-2=4+(1/3)a\n\nsubtract (1/3)a from (3/2)a and add two to both sides:\n(3/2-1/3)a=4+2\n(9/6-2/6)a=6\n(7/6)a=6\n\nnow solve and what do you get? :)
842934
Ok, first you must get the -8 and the x/-2 out of parentheses, so u just multiply them both by one. So then is -8 * x / -2. First, you do -8 * x, which is -8x. The problem then becomes -8x / 2. In this step, just don't mind the negative or the x and just put them back in at the end. So you're really just doing 8 / 2, which is 4. You must put the - and the x back in, so the answer is -4x. Ta Da!
191567
Please use parentheses to explain the problem more effectively. Assuming the problem as,\n\n(x-1)/2x + x/(x+3)\n\nThe solution can be,\n\n(3x^2+2x-3)/(2x^2+6)
520168
Try this:\n\nhttp://www.algebra.com/
896553
4x^2+8x-x-2\n(4x-1)(x+2)
780960
I also have difficulty typing this thypes of math problems.  First, simplify the ax^2+bx+c parts in the first half to (x+c)(x+c) looking things first.  So...\n\n[(x-3)(x-1)]} is the first part.  With the bottom of the second half, divide by 3, so that you get 3(x+1).  Now line everything up like so...\n\n(x+1)(x+3)(x-3)\n_____________\n3(x-3)(x-1)(x+1)\n\nNow do the cancellations and you get the right answer...\n(x+3)\n____\n3\n\nSorry for not working to convention, but it was easier to type it that way.
912865
bring the a's to one side... u then get 9a = 27 ... divide and answer is a = 3
626693
ax - b = -c\nax = b - c\nx = (b - c) / a.
503153
(x+7)(x+3)\n(x-11)(x+2)\nforget this one.  its AC factoring though\nthis ones AC too
890098
I enjoy algebra, and this is easily my favorite type of problem.  So I hope you, too, one day enjoy them for their inner beauty.\n\nFor now, let's tackle this "factoring problem."  We may wish to use the method of "synthetic division" which is arguably easier than "long division of polynomials."\n\nThink back to grade school where you learned that 6 = 2 * 3.  This simple statement enabled you to 'deconstruct' the number 6 as either "two times three" or "three times two".  Even if you forgot one of the 'factors' you could conjure it up again quickly by asking yourself, "What is six divided by two?" for example.\n\nIn problem (a), you are essentially asking, "What are the coefficients of P(x) if the only factor we know is (x-1)?"\n\nLet's set up synthetic division (see link below), by placing all the coefficients of P(x), and using 1 (the zero implied by (x-1)) as the divisor:\n\n1 ||  a  b  c  d\n___________\n.....a\n\n1 ||  a  b  c  d\n_____a_____\n.....a (a+b)\n\n1 ||  a  b        c       d\n_____a___(a+b)____\n.....a (a+b) (a+b+c)\n\n1 ||  a  b        c                d\n_____a___(a+b)___(a+b+c)_\n.....a (a+b) (a+b+c) (a+b+c+d)\n\nNote that this last term must equal 0 since it is assumed that (x-1) is a factor ... thus, the "remainder" term must equal zero!\n\nThere's the result you were after:  a+b+c+d = 0\n\nHope this helps you tackle the others!
87883
First write y as a product instead of a quotient:\n\n(2+x)*(2-3x)^(-1)\n\nNow you should be able to do this. \nUse 2+x as your f and (2-3x)^(-1) as your g. \nYou will have to use the chain rule to find g'.