factor the following polynomials completely  -3x^2-10s+8 now would i find the middle factor...?
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)
968026
I got   3(x² - 5x + 4) or 3( x - 4 )( x - 1 )\n\nLet me show you my work...\n\n3( x - 2 )² - 3(x - 2) - 6\n\nSimplify\n\n3( x -2 )( x - 2 ) - 3( x- 2 ) - 6\n\nDistribute\n\n3( x² - 4x + 4) - 3x + 6 - 6\n\nDistribute and Simplify\n\n3x² - 12x + 12 - 3x\n\nSimplify\n\n3x² - 15x + 12\n\nYou can now simplify by factoring out 3\n\n3(x² - 5x + 4)\nor\n3( x - 4 )( x - 1 )
423594
(3x^3-7x^2+x-20)/(x-3) = 3x^2+2x+7 with a remainder of 1\n         \n         3x^2+2x+7 \n     _______________\nx-3 | 3x^3-7x^2+ x-20\n      -3x^3+9x^2\n    __________\n          2x^2+ x\n         -2x^2+6x\n         ________\n              7x - 20\n             -7x + 21\n             ________\n                   1\n\nI tried to show what this would look like as a division problem but it keeps jumbling up.
854837
Show it's a zero just means you sub x = -3 in the polynomial and get the result zero.  If that happens, you can divide the polynomial by \nx -(-3),  i.e. x + 3, and then find the zeros of the remaining expression.  Want to do it yourself?  Good.  Then don't scroll down to see what I've done.\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n-3      2     5     -6    -9\n\n              -6       3      9\n\n         2    -1     -3      0\n\nP(x) = (x + 3)(2x^2 - x - 3)\n       =  (x + 3)(x + 1)(2x - 3)\n\nZeros are -3, -1, 3/2
503153
(x+7)(x+3)\n(x-11)(x+2)\nforget this one.  its AC factoring though\nthis ones AC too
967957
There really isn't any way to do this other than guess and check. If you find that you substitute some value in and find f(a) = 0, then (x-a) will be a factor. You can repeat that a few times until you have found some roots, and then do a long division to something which you *can* solve by hand (a quadratic or linear).\nNote also that you only need to check things which are factors of the constant term, and there aren't too many.\nIn the first case, you will find that substituting in -2, -1 and 3 will all give 0, but you may have trouble finding the last root. So you will have to do some long divisions (divide by (x+2)(x+1)(x-3)), and you will end up finding that the -1 was a repeated root - so the answer is (x+2)(x+1)^2(x-3).\n\nSame thing for the second: you will find -1 as one root. You can then do a long division and find -1 is repeated, and again, and again, and again. The actual factorisation is (x+1)^5 - you may have noticed this if you knew much about pascals triangle and recognised the coefficients are an entire row - 1,5,10,10,5,1.
219588
3x^3 + 2x^2 + x - 6\n= 3x^3 + 5x^2 - 3x^2 + 6x - 5x - 6\n= x(3x^2 + 5x + 6) - 1(3x^2 + 5x + 6)\n= (3x^2 + 5x + 6)(x - 1) \n\nThe factors are : (3x^2 + 5x + 6) and (x - 1) \n\nAlternatively, using polynomial division : \n(3x^3 + 2x^2 + x - 6) divided by (x-1) yeilds \n(3x^2 + 5x + 6)
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
223322
3x*(x^2-8xy+16y^2) Same factor\n3x*(x-4y)^2 Perfect squares.
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)
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)
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
821200
(x²+4x+4) / (2x-3) / (x²-4) / (2x²-7x+6)\n\nYou have 4 terms and three divisions, so it's a little unclear if you mean (a/b) / (c/d) or what.  I will try to help to the extent I can.\n\nLet's assume you mean:\n\n[ (x²+4x+4) / (2x-3) ] / [ (x²-4) / (2x²-7x+6) ]\n\nFactor your x² equations to see if you can cancel terms:\n\nx²+4x+4 = (x+2)(x+2)\nx²-4 = (x+2)(x-2)\n2x²-7x+6 = (2x-3)(x-2)\n\nWoohoo - things are looking good!  Now we can rewrite our original problem using these factors:\n\n[ (x²+4x+4) / (2x-3) ] / [ (x²-4) / (2x²-7x+6) ] = \n\n[ (x+2)(x+2) / (2x-3) ] / [ (x+2)(x-2) / (2x-3)(x-2) ] =\n \n[ (x+2)(x+2) / (2x-3) ] * [ (2x-3)(x-2) / (x+2)(x-2) ] =\n\nnotice we can cancel out three terms: (2x-3), (x-2) and (x+2)\n\n[ (x+2)(x+2) / (2x-3) ] * [ (2x-3)(x-2) / (x+2)(x-2) ] =\n\n(x+2)\n\nHope that helped!\n\n~ ♥ ~
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
770521
When your first term is x², you can look at the other two coefficients -- in this case -8 and +15.\n\nTo factor, look to see what two numbers when multiplied together will give you your last coefficient (+15) and when added together will give you your middle coefficient (-8).\n\nSo in your case, we see that we need two NEGATIVE numbers because 15 is positive and only two negs or two positives can give you a positive when multiplied.\n\nWe know that the two must be negative, because we need to get a negative 8 when we add them together.\n\nSo, we can see that +15 = -5 * -3\nand -8 = -5 - 3\n\nSo, our factorization is (x-5)(x-3).\n\nNotice we plugged in our -5 and -3 into (x__)(x__).\n\nHope that helped somewhat!\n\n.
846363
3x² + 6 + 10 - (3x² - 8x - 5)\n\n3x² + 6x + 10 - 3x² + 8x + 5\n\n3x² - 3x² + 6x + 8x + 10 + 5\n\n0 + 14x + 15\n\n14x + 15\n\nThe answer is 14x + 15\n\n- - - - - - -s-
408031
First, factor the polynomial using the Rational Roots Theorem. All possible factors of -36 are (+or-) 1, 2, 3, 4, 6, 9, 12, 18, 36.  Use synthetic division for each, until you find one root.  In my solution I got x - 4 as a root and the other x^3 + x^2 + 9x + 9.  Continue with the factoring.  Here I got x + 1 as a factor and the other x² + 9. Therefore, the factorization is:\nP(x) = (x - 4)(x + 1)(x² + 9)\n\nThe zeroes are all the roots when P(x) = 0\nTherefore,\n(x - 4)(x + 1)(x² + 9) = 0\nx = 4\nx = -1\nx² = -9 or \nx = 3i or -3i\n\n^_^