14x^2-20x+60 so i am looking for the numbers that add up to the middle number which is 20 in order to factor?
539132
both roots are complex numbers
631080
You can noy factor it out... it makes it easier to figure out when you put it into lowest terms... so it would be \n\n   x^2+7x-25..... so you cant factor it.
780226
lets take the common multiple 210\n15,35,21 and 2 are the factors\nI was thinking of 105 but only 15.35.21 come to my mind
769323
6 - 12x\n-or-\n-12x + 6\n\n-or-\n\n6(1-2x)\n\n...although that depends on whether you want the simplest expanded form or the simplest factored form.
959197
I have factored it please refer to.\n\nFrom this solution can be found
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!
149556
4x^3+36x^2+32x----using common factor which in this case would be 4x so, 4x(x^2+9x+8) and then u use guessing method \nand the complete factorization would be 4x(x+8)(x+1).
402740
if you mean 3 , 2 ,  6 then it is the answer :\n(3 * 6) +2 = 20 .
1091081
64x^6 - 576x^5/y^2 + 2160x^4/y^4 - 4320x^3/y^6 + 4860x^2/y^8\n- 2916x/y^10 + 729/y^12\nIt's the same as mine except I put the y's in the denominator...but that doesn't matter.
310788
You need to factor everything, then cancel common factors.\n4x+8 factors as 4(x+2)\n6x-3x^2 factors as 3(2-x)\n3x^2-4x-4 factors as (3x+2)(x-2)\n9x^2-4 factors as (3x+2)(3x-2)\n\nCancel the (3x+2)'s, and the (x-2) can cancel with the (2-x) leaving a factor of -1.  The factors that are left are:\n[-4(x+2)]/[3(3x-2)].
842037
We need to factor the equation down. This involves finding the roots of the equation. This will look like:\n\n(ax+b)(cx+d) = 12x^2-31x+20 = 0\n\nBecause the final term is positive and the middle term is negative, my first hunch is to have b = -4 and d = -5. By doing so, the equation looks like:\n\n(ax-4)(cx-5) = 12x^2-31x+20 = 0\n\nThe factors needed for 12x^2 could be 2 and 6 or 3 and 4. We should test each combination to see that they work out. The factoring ends up looking like this:\n\n(3x-4)(4x-5) = 0\n\nNow all that remains is to solve for x in each set of parentheses. If the first term is zero, then\n\n3x-4 = 0\nx = 4/3\n\nIf the second is zero, then\n\n4x-5 = 0\nx = 5/4\n\nThus, x = 5/4, 4/3
485469
product of 8 and -3  = -24\nfactors for -24 are 1)-12 ,2\n                    2) 12, -2\n                    3) 4 , -6\n                    4) -4 , 6\n                    5) 8, -3\n                    6) 3, -8\n\nOf the above only the factors 1), that is -12, 2 when added give -10.\n\nTherefore we would take those factors. And split the middle term to get the following:\n\n               -3x^2 -12x + 2x + 8\n\nNow factorize to get:   -3x( x + 4 ) + 2 ( x + 4 )\n                      = (-3x + 2)( x + 4)
260457
20 - - 4 = 24\nminus and minus = plus\nie\n20 - - 4 = 20 + 4 = 24
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
6182
You can use what is called the Master Product Method to factor this and other quadratic trinomials.\n\nFor something of the form ax^2 + bx + c, the master product is a*c.  In this example, a is 2, b is -5 and c is 2, so the master product is 2*2 = 4.  Now you must find two numbers which multiply to give you the master product of 4, but add to give the middle coefficient of -5.\n\nThis requires a bit of trial and error, but you should be able to quickly discover that the numbers -1 and -4 will work since -1*-4 = 4 and -1+-4 = -5.  Now you rewrite the the entire trinomial replacing the middle term -5x with the equivalent -1x - 4x:\n\n    2x^2 - 5x + 2\n= 2x^2 - 1x - 4x + 2\n\nWe factor by grouping the first pair and last pair together and pulling out what's common to each pair:\n\n= (2x^2 - 1x) + (-4x + 2)\n= x(2x - 1) -2(2x - 1)\n\nNotice that (2x-1) appears in the factorization of both pairs.  We finish by factoring out the (2x-1) to get\n\n= (x - 2)(2x -1)
948048
This seems kinda beyondf algebra two, finding all whole number (positive integers except zero) values satisfying this.\n\nTry this.\n\nx^2 - y^2 = 48, so factoring gives (x + y)(x - y) = 48.  Since x and y are positive integers, you must have x > y.  So x + y and x - y are also positive integers, and moreover x + y is greater than x - y.\n\n\nNow factor 48 every possible way into two integers:\n\n48 * 1\n24 * 2\n16 * 3\n12 * 4\n8 * 6\n\nYou can set up five systems of equations, one for each factor pair.  For example, the first system would be \nx + y = 48\nx - y = 1\n\nThe soltuion to this is x = 48.5 and y = 0.5.  Since these aren't integers, discard them.\n\nIn order to be integers, the parity (evenness/oddness) of the two factors must be the same.  So you only need to consider 24*2, 12*4, and 8*6.  the system from 24*2 would give x = 13 and y =11.  The system from 12*4 would give x = 8 and y = 4 (your solution).  The system from 8*6 would give x = 7 and y = 1.
964841
Figure it out - when you divide 60 by each of those numbers, which one leaves a remainder?