Use a special product formula to factor the perfect square trinomial.?
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.
204269
isolate one of the radicals, then square both sides.\nhere you get 2x+3=5x-6.......3x=9.......x=3\n\nworst case is sqrt(something) + something = sqrt(something)\n\nhere you'd isolate a radical, square both sides, clean up the equation, isolate the remaining radical, square both sides, and finish.  AND CHECK YOUR ANSWERS.  when you square both sides, "extraneous" solutions sneak in that might have you, back in the original equation, taking the square root of something negative.\n\nexample:  sqrt(3x+3) + 1 = sqrt(5x+6)\n....... (3x+3)+2*sqrt(3x+3)+1=(5x+6) ..... left side is a binomial\n.........2*sqrt(3x+3) = 2x+2\n...........sqrt(3x+3) = x+1\n...............(3x+3) = x^2+2x+1 ....... right side is a binomial\n....................0 = x^2-x-2 ......... and it factors!\n....................0 = (x-2)(x+1)\n....................0 = (x-2) so x=2\nor .................0 = (x+1) so x=-1\n\nnow check: sqrt(3*2+3)+1 =? sqrt(5*2+6)\n.............sqrt(9) + 1 =? sqrt(16)\n...................3 + 1 = 4.............yes.\n\n..........sqrt(3*-1+3)+1 = sqrt(5*-1+6)\n...............sqrt(0)+1 = sqrt(1)\n.......................1 = 1 ............yes\n\nsigh.  when I wrote it the equal signs lined up.
773695
Look in your text book under "factoring polynomials"... or you could try this web site that explains this fairly thoroughly...\n\nhttp://faculty.ed.umuc.edu/~swalsh/Math%20Articles/Factoring.html
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.
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)
10610
X^2-9X+18\n\nShort and simple just factor it FOIL
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!
961776
That is not the quadratic formula it is -b plus or minus, b squared - 4(ac) all over 2a, you would best to just look up the quadratic formula to understand it, you can also use factorization, or completeing the square, but must have a coefficient of one.  For larger pol nomials there are other ways, but just best to google it or yahoo it considering the site i am on.
893470
x^2-4x-5 = (x+1)(x-5) if that helps any.  I just factored the trinomial into two binomials.  A graphing calculator would help you solve this in no time.  I have my own homework to do, so I'm not gonna do this for you, but I hope that that will get you started.  Good luck.
917477
where else?\n\nhttp://www.factoring-polynomials.com/
444843
It's a perfect square trinomial.  Take the square root of the first term and the square root of the last.  If you multiply them and then double it, and it equals the middle term, then it's a perfect square trinomial.\n\nIn this case, it factors into (x/3 + 6)^2.  Using FOIL, you'll see that it expands into your original expression.
958614
#1.)-----------------------------------\n  \n25b^ -20by + 4y^\n\nMultiply 1st and 3rd term together to get 100b^y^.\n\nFactor 100b^y^ into two factors whose sum = the middle term of -20by.\n\n1by   X 100by\n2by   X 50by\n4by   X 25by\n5by   X 20by\n10by X 10 by   ....................BINGO!\n\n-10by + (-10by) = - 20by   &   (-10by)(-10by) = 100by\n\nThis is a really kewl factor by grouping trick, I LOVE it!\n\nBring down the first term of: (25b^)\n\nNext, write the SMALLER of the two factors 1st.  Followed by the LARGER of the two factors.  In this case, they're both the same, so it doesn't matter which one goes first: (-10by)\n\nFinally, bring down the third term of: (+4y^)\n\n25b^ -10by -10by + 4y^\n\nNow, lets factor by grouping! \nFactor out the GCFs.\nGCF#1 = 5b\nGCF#2 = -2y\n\n(25b^ -10by) + (-10by +4y^)\n5b(5b - 2y) -2y(5b - 2y)\n\nGCF#3 = (5b - 2y)\n\n(5b - 2y)(5b - 2y)\n\nSimplifies further as: (5b - 2y)^   ....Whatdayaknow?! \n.......................................................................................................It's a Perfect Square Trinomial.\n\n#2.)------------------------------------------------------------------\n\n100 - 9b^\n\nAh, the difference of two perfect squares.\n\n---------------------------------------------\n\nFORMULA:\na^ - b^ = (a-b)(a+b)\n\n---------------------------------\n\nSOLUTION:\n100 - 9b^\n(10)^ - (3b)^\n\n-----------------------\n\nANSWER:\n(10-3b)(10+3b)
382213
A perfect square trinomial is in the form\nm² + 2am + a²\n\nSince you know 2a is -8, a must be -4\nSince n = a², n = (-4)² = 16
773858
If you work out the square (of -3) and the multiplications and divisions to find the value that you need to take the square root of, you will find that the number is 121. And you should be able to calculate the square root of 121.\n\nIt would be helpful to you if you developed your ability to factor trinomials like this, so that you could solve problems like this by factoring. I used to use the quadratic formula all the time, and in real-world problems it's often the only practical way to get an answer. But for school problems, many, many times you can do it much faster by factoring.
1053403
Write it as:\n2x^2-4x+3x-6\n=2x[x-2]+3[x-2]\n=[2x+3][x-2]\nSimple middle-term factorisation!!!
1002101
factor 9x^2y^2-9y^2\n9y^2(x^2-1)\n9y^2(x+1)(x-1)
766790
1. 5m^2 - 15mp - 2mr - 6pr <-- Impossible to factor by grouping. It needs to be either, exactly one of the signs needs to be reversed.\n\n2. 6-3x-2y+xy\n(6 - 3x) - (2y - xy)\n3(2 - x) - y(2 - x)\n(3 - y)(2 - x)