polynomials and factoring???
924243
(q - 6h)(g - 6h) \n= q(g - 6h) - 6h(g - 6h)\n= gq - 6hq - 6gh + 36h² (-6h x -6h = +36hh = +36h²)\n\n8q²j - 12g³hj \n= 2 x 2 x 2 x q x q x j - 2 x 2 x 3 x g x g x g x h x j\n= 2 x 2 x j x (2 x q x q - 3 x g x g x g x h)\n= 4j(2q² - 3g³h)  (The common factors in each part are the multiplier, the remaining factors are what are left inside the parentheses ... and always remember that 1 is always a factor of any number .. see example below)\n\n5g²h - 10g³h²\n= 5 x g x g x h - 2 x 5 x g x g x g x h x h\n= 5 x g x g x h x (1 - 2 x g x h)\n= 5g²h(1 - 2gh) (because 5g²h x 1 = 5g²h)
32760
You can use the AC Method to answer this question:\n\nFirst, rewrite the expression so it's in the form Ax^2 + Bx + C where A,B, and C are constants (numbers)\n\n16x^2 + 8x - 15\n\nThere's no common factors we could take out first, so...\n\nMultiply A x C, which in this case is 16 x -15 or -240.  Then look for factors of this number that add up to B.  You're looking for two numbers that fill in the blanks below:\n__ x __ = -240\n__ + __ = 8\n\nPrime facorization will help give you some ideas, or make an ordered list of numbers that multiply to -240.  \n240 and -1\n120 and -2\n80 and -3\n60 and -4\n48 and -5\n40 and -6\n30 and -8\n24 and -10\n20 and -12\n\nSee how the last pair of numbers satisfies the conditions?\n\nRewrite your polynomial, substituting the 2 values in for B to have 4 terms:\n\n16x^2 + 20x - 12x -15\n\nThen factor by grouping\n(16x^2 - 12x) + (20x - 15)\n4x (4x - 3) + 5 (4x - 3)\n(4x + 5)(4x - 3)\n\nThis method works for most polynomials of the form Ax^2 + By + C!  \n\nYou can check your answer of (4x + 5)(4x - 3) by using the FOIL method and see that it does work.\n\nGood luck!
835153
.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
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)
930643
factor:\n\n(2x^2+x-6)/(2x+4)=\n\n(2x-3)(x+2) / 2(x+2)\n\nthe (x+2)s cancel out leaving\n\n(2x-3)/2
1177677
shouldn't it be x^2??? if it IS x-squared, then sure, you could use the factorization method. \n\nit goes like this:  (6x + 1) (x - 1) = 0\n\ntherefore, x = -1/6 , 1
614345
(x+10)(x-10)\n(x-4)(x²+4x+16)\n(x+w)(x²-wx+w²)\n\nUnfortunately you just have to memorize the formulas.  There is no other way to figure out the factoring of the cubic expressions.
221826
1) \nxy - x + 4y - 4\nx(y-1) + 4(y-1)\n(x+4)(y-1)\n\n2)note: x^3 means x cubed; x^2 means x squared\nx^3 -6x^2 + 2x - 12\nx^2(x-6) + 2(x-6)\n(x^2 + 2)(x-6)\n\n3)\n6tx + r - 3t - 2xr\n3t(2x - 1) - r(2x - 1)\n(3t - r)(2x - 1)\n\n4)Are you sure it wasn't -x^2 - 2x - xy - 2y?\n-x^2 - 2x - xy - 2y\n-x(x + 2) - y(x + 2)\n(-x - y)(x + 2)\n\nTo "factor by grouping," you must split the original polynomial that you are given into two different binomials by "grouping" them together.  This is done with perenthesis in the problems above.  Then, you factor out the "greatest common factor" from each of the binomials.  After factoring, the two quantities still within the perenthesis should be the same.  You could call these quantities the "common binomial."  Lastly, you factor out the common binomial to get the final answer.
594239
ok so you have p^2 + p, you can factor out p from both parts of the expression, it will end up like this: p(p + 1).  Hope that works.
294579
First multiply p times the second polynomial.\np(p^4 - 2p^3 + 3) = p^5 - 2p^4 + 3p\nthen multiply 2 times the second polynomial.\n2(p^4 - 2p^3 + 3) = 2p^4 - 4p^3 + 6\nNow add these two together.\n\n(p^5 - 2p^4 + 3p) + (2p^4 - 4p^3 + 6)\n= p^5 - 4p^3 + 3p + 6
781910
not possible to factor easily ... try the quadratic formula.
498372
I think u are talkin abt this method....\n\n\nif u hav a cubic like  x^3 + 2x^2 - 2x - 1 = 0\n\nto find out whether (x-1) is a factor....\n\nfirst write the coeff of the eq in this way n write the factor ( here it iz 1 since x-1=0 or x=1) on the left hand side Like this\n\n{one}  1     2     -2     -1\n--------  0     1      3      1\n_______________________________\n{sum}  1     3      1      0\n\n\nHere what i hav done is write the 0 under the coeff of x^3\nnow add that coeff n the zero\nu get 1...\nnow multiply this one wid the number written on the left hand side(here it is also one)\nwite what u get under the coeff of x^2....\nas above add the new no and the coeff of x^2\nin a similar way multiply one and the sum u got n write it under the coeff of x...\nadd the no with coeff of x n then write it below....\nfinally u multiply the sum wid one and write it under the constant term.... n then add...\nif u get 0 as the sum then whatever no u hav taken in da LHS is on of the root.... here we get 0 in the last case... so (x-1) is a fctor.....\n\nnow this eq can b written as \n\n    (x-1) * (1*x^2 + 3 * x + 1) = 0\nhere 1 , 3 and 1 of the quadratic are the sum u got in the above method......
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)
570705
Factoring out X gives:\nX(-X^2 + X + 6)=0\nUsing the quadratic formula to factor the terms inside the brackets gives:\n-X^2 + X + 6 = (X+2)(-X+3)\nSo we have:\nX(X+2)(-X+3)=0 and the zeros are X=0, X=-2, X=3
607601
I think you set the equation to zero then you have to factor it.  Go to the link below if you wanna know how to do it. OR look your your math book.
168995
plynomial expression could be factorized as the product of \n(x-r) where r is a root (a zero) of the polynomial soyou must get the roots first\nMany numerical methods exists
824178
You typed exponents just fine!\n\nx^3 - x^2 -x + 7 divided by x+1 is a "long division" type problem rather than factoring because x+1 doesn't go evenly into the polynomial\n\nThe answer is x^2 -2x + 1 + 6/x+1