So far your help has got me a 90% and i thank everyone, but now i need more help with factoring.
213146
Factoring is the "opposite" of FOILing.\n\nIf you had x^2 (which is x squared)\nx^2 + 5x + 6\n\nYou ask yourself:  What multiplies to 6?  (Write all the choices including negatives)\nThen, out of those choices, which one (and only one will work) adds to 5.  \n\nSo... 2 x 3 = 6  and -2 x -3 = 6\n2 + 3 = 5\n-2 + -3 = -5 \n\nSo, 2 and 3 add to 5 and mult. to 6.\n\nThen you write it in parenthesis keeping the signs.\n\nYou get (x + 2)(x + 3)\n\nHope that helps.  Good luck.
442605
.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
303295
I would be happy to help.
293574
the first step is basically to factor 120175.  the obvious factor to do first is 5.  this gives us 5*24035.  then factoring another 5 gives us 5*5*4807.  since 4807 obviously isn't divisible by 5 or 2, try 3.  doesn't work - so keep on trying the next biggest prime number (1,2,3,5,7,11,13,17...)  turns out that 4807 is divisible by 11.  this gives us 5*5*11*437.  keep on going and you get 5*5*11*19*23.  just checking, yes - 5*5*11*19*23=120175.\n\nnow converting these factors into letters, you get the available letters EEKSW.  this can be rearranged to be WEEKS.  however, remember that 5*5*11*19*23 can also be written as 25*11*19*23, since this also equals 120175.  this means the word could also have KSWY.  but i don't think you can rearrange those letters to make a word.\n\nso therefore the english word encoded as 120175 is WEEKS.  basically, you factored the number, converted them into letters, and solved the anagram.\n\n[[edit]]\n\ninuyasha's right - there could be an infinite number of "a"s in there too since 5*5*11*19*23 = 5*5*11*19*23*1*1*1*...\n\nhowever, this gives us WEEKS+AAA... so using the online anagram solver, found at http://wordsmith.org/anagram/ , there are no words that have "WEEKS" in them and "A"s.  therefore, the answer is still WEEKS.
445221
I would factor each quadratic and then see if I could somehow cancel some stuff out.\n\nx² - x - 6 = (x-3)(x+2)\ny² + 3y - 10 = (y+5)(y-2)\ny² - 7y + 10 = (y+5)(y+2)\n3x² + x - 10 = (3x-5)(x+2)\nx² - 2x - 3 = (x-3)(x+1)\n9x² - 25 = (3x+5)(3x-5)\n\nRewrite your problem with these factors and you will see that you can cancel out a lot of these terms!
534479
#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)
492617
wow its been a long time.. i use a calculator to do that now.. but here goes one thing i remember...\n\nlets say we have a polynomial like this\n\nx^2 -3x-10  and we need to factor it..\n\nwe take a look at what factors of -10 we could use that would add up to -3  in this case there are two way to get -10 (-2*5) or (2*-5)  if we add the two sets together one gives us -3 and the other gives us 3 so we can throw out the -2*5 terms.\n\nSo now we lay out the factors.... we need two x's to get x^2 so we need two x's in the first terms (remember FOIL?)\n\nlay the terms out so they give us the correct values\n\n(x+2)(x-5)  now we can use FOIL (firs, outer, inner, last) to check ourselves...\nX*X = x^2\nX*-5 = -5X\n2*x =2x\n2*-5=-10\n\nso we get [x^2] +[-5x] +[2x] +[-10]\nor x^2 -5x+2x-10\ncombine like terms now\n-5x+2x =-3x\n\nwhich gives us\nx^2-3x-10\n\nhope that helps
306199
8x^2y^3+5y^2 when "FACTORED COMPLETELY"=\n\ny^2 (8yx^2+5). \n\nThis should be right. My dad is a complete genius and he did this in like 2 seconds. I double checked it on my calculator too by plugging in numbers, so I can assure you he's right.\n\nHope this helps! Have a nice day :)
234031
It's called factoring by grouping. Take a factor out of two of the four terms (and the same for the other two) so they share a common factor.\n(2x^3 - x^2) - (18x - 9)\nx^2(2x - 1) - 9(2x - 1)\n(x^2 - 9)*(2x - 1)\n(x + 3)*(x - 3)*(2x - 1)\n(ungkog WAS correct)
419119
notice that (4x+36) has a factor of (x+9) and set up on fraction with a denomicator of (x+9)
476810
It's obvious that 3 can be factored out of each term immediately.  That leaves us with: \n\n3(2x^2+xy-3y^2), \n\nwhich can be factored much more easily than the original expression.\n\n(2x^2+xy-3y^2) factors into (x-y)(2x+3y).  \n\nSo we are left with the following complete factorization for the expression:\n\n3(x-y)(2x+3y)
488377
i still gota bunch of math<:( grrr....
364719
Well, looks like you have the answers.  My work here is done.
76248
Yeah, I know what you mean. Read my answer to your first question, too, b/c maybe another teacher might be a better match for you as a tutor.  My daughter is having the same problem as you right now, her teacher is so boring she is killing the class slowly and surely. Used to be my son could tutor her, but he is in college now. I can't help b/c you kids are doing math I never ever heard of.  She tells me that you can go to some websites that give math help, if you google the math term you don't understand.  Good luck and don't give up, please?
436597
Obviously 1 isn't worth trying because 90 is so much bigger than the other terms, but you should check that\nP(2) = 0, and so\nP(x) = (x - 2)(x^3 + x^2 - 21x - 45)\n[You know how to do long division to get this?]\nSub -3 in the second factor and find\nP(-3) = 0\nand so x + 3 is a factor, and then\nP(x) = (x - 2)(x + 3)(x^2 -2x  -15)\nand you should be able to finish it from there.
381414
ok re arranged the second question so that the 6 is factored out so its x= such and such then sub the x equals such and such into the first equation and solve then do the same thing to Y in the second equation
580396
the 3rd one is correct, but I get -(3+b)/b+2 and 2 for the first 2