How is this problem factored?
954091
-4t^2+36-72\nFactor-2t+12\n +2t-4t^2+24\n-6+12-72\nIt is (2t-6)(-2t+12)\n\nFirst you put the first trinomial in the upper left corner of the box (-4t^2).\nThen you put the last trinomial in the bottom right corner of the box(-72).\nYou multiple (-4t^2 & -72) then you make a list of factors that will equal that number. In this case they are 12 & 24.\nYou know that one number must be negative because -4t^2. Fill in the info in the box and multiple them out. \nIf you do this you can see that it equals the trinomial.\nAdd down the left side of the box to get the first set of numbers.\nThen add across the box to get the second of numbers.\nYou can see how I did this.\n\nYou can email me if this doesn't make sense.
447887
You probably mean\np(x) = 2x^4 +5x^3 – 3x^2 -2x -6\nThe answer is e) Both b,d\nbecause when you multiply the factors you should get the original expression and (x+3)(2x^3 –x^2 -2)= 2x^4 +5x^3 – 3x^2 -2x -6\nThis was quickly seen because you need x^4 which is only obtainable by x multiplied by x^3 in your problem :)\nHope this helps
777394
factor out 4b\n\n4b(b-2)= 0\n\nb = 2, 0\n\ngoodluck ^_^
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.
247403
12.\n\nEverytime you see the question...  What number is both a factor and a muliple of X, X will be the answer.
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
218359
may be you dont know the system..\nok let me help you...\n2x^2-9x-5 this can be factorised using middle tram factor..\nlet me write it in general from..\n(mX^2+nX+c); first find m*c let k=m*c\nand then factorise k in such from that a*b = k;\nand a+b = n;\nat last rewrite the problem as...\nmX^2+aX+bX+c and solve\n\nin your problem m = 2; n = -9; c = -5;\nk = 2*(-5) = -10;\nfactorising 10: -10 = 10*(-1) = 2*(-5)...\nhere -10+1 = -9 = n\nso a=-10, b=1\nrewriing \n  2x^2-9x-5\n= 2x^2-10x+x-5\n= 2x(x-5)+1(x-5)\n= (x-5)(2x+1)
807869
the prime factorization can be figured out by repeatedly dividing the number by a prime number.\n for example 12\n 12 divided by 2 = 6,   6 divided by 2= 3,  3 divided by 3= 1\n the answer is 1.\ni hope this was helpful.
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
777032
Since there are too many numbers to list, here's a program to do it for you:\nhttp://wims.unice.fr/wims/en_tool~algebra~factor.html
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!
930308
x=[2+/-rt(4+60)]/2\n=[2+/-8]/2\n=-3 or 5\nso the factors are\n(x+3)(x-5)
208169
Factoring, I guess, but it really depends on the type of polynomial question. Can you give a sample?
220199
The problem factors to\n\n(x - 2)(x^2 + 1)(x^2 + 1) or (x - 2)(x^2 + 1)^2\n\nx = 2, i, -i\n\nthe only real zero is 2.
807151
you can factor by using this:\n(y + 4)( y + 4)=0\n\nnow you have\ny + 4 = 0\n\nsubtract 4 from each side:\ny + 4 = 0 \n   - 4    -4\nyou'll end up with:\ny = -4\n\nPS please award best answer
781910
not possible to factor easily ... try the quadratic formula.
805233
Can you factor the numerator?\n\nI guess not, but UMR math major ignores your denominator in the original problem.