Factoring Quadratic Trinomials?
928591
x^2-6xy+8y^2\nx^2-6xy+9y^2-9y^2+8y^2\n(x-3y)^2-y^2\n(x-3y+y)(x-3y-y)\n(x-2y)(x-4y)\n\naliter\nx^2-4xy-2xy+8y^2\nx(x-4y)-2y(x-4y)\n(x-4y)(x-2y)
503153
(x+7)(x+3)\n(x-11)(x+2)\nforget this one.  its AC factoring though\nthis ones AC too
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)
684474
You have to factor each part first, then cross off (cancel) identical factors from the numerator and denominator of either expression.  If the factors are exact opposites, like x-8 and 8-x, cancel and put a -1 in the numerator.\n\n\nIn your example only the first expression is factorable:\n(x-8)(x+3)/(2x)(3+x) times 5x^2/(8-x)\n\nThe (x-8) cancels the (8-x) to become -1, then (x+3) cancels the (3+x), and the x from the (2x) cancels one of the x's from (5x^2)\n \nAnswer: -5x over 2
826369
I have found sites like the following helpful:\n\nhttp://www.algebra.com/\nhttp://www.saab.org/mathdrills/factor.cgi\nhttp://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=algebra&s2=factor&s3=basic\n\nThe last one is a solution program, but it can be helpful to check your answers.
1095294
5) find the factors\nsquareroot of x * x * x * x * x * x /  7*7*5*5*2*2\nthe factors can come out of the square root by pairs only\nyou have 3 pairs of x's = x^3\nyou have 1 pair of 7's so it comes out of the square root as a 7\nx^3/7*5*2\nx^3/70 is it's simplified term\n\n__________________\nif u want all the correct answers then that would tarenirator's answers
836052
You can start by factorizing 3x^4 from everything\nm(x) = 3x^4 (2x^3 + 5x^2 - 3x - 3)\n\nI think there's something wrong with your problem, none of the possible roots given by the Ruffini's rule are working. Perhaps, if instead of 6x^7 + 15x^6 - 9x^5 - 9x^4 it was 15x^7 + 15x^6 - 9x^5 - 9x^4, then it could be factorized very nicely in\nm(x) = 3x^4 (x+1)(5x^2-3) \nor if instead you wrote 6x^7 + 6x^6 - 9x^5 - 9x^4 then it'd factorized to\nm(x) = 3x^4 (x+1)(2x^2-3)
439883
Use the following four steps to solve a quadratic equation, in this order:\n\nStep 1: Guess the answer\n\nIf x^2=0, can you guess the answer?\nIf x^2=1, you can guess, but remember that there may be two answers. (1 and -1)\n\nYou can tell if you are guessing right by plugging back into the original equation. If it worked, you're done.\n\nIf you can't guess in 30 seconds, you are done with Step 1. Go to Step 2.\n\n\nStep 2: Try factoring\n\nx^2 + 4x + 4 = (x+2)(x+2)\n\nRemember that the constant term of your factors has to multiply to the constant term of the quadratic. \n\nFor example, if you have x squared plus something plus 6, remember that your constant terms must be multiples of 6. \n\nTry:\n\n(x + 1)(x + 6)\n(x - 1)(x - 6)\n(x + 2)(x + 3)\n(x - 2)(x - 3)\n\nYou just ran out of factors of 6. If this didn't work, go to step 3.\n\n\nStep 3 -- Complete the Square\n\nOne important thing to remember about completing the square is that it does exactly the same thing as the quadratic equation. Pick either Step 3 or Step 4 as your last step. Guessing and factoring don't always work because you can't always guess and you can't always factor. But completing the square and using the quadratic formula will always find the answer if there is one.\n\nSuppose you have \n\nx^2 - 6x - 16 = 0\n\nHere's how to complete the square:\n\n#1. Move the constant (whatever it is) to the right side\n\nx^2 - 6x = 16\n\n#2. Find half of the coeficient of the x term and add the square of that number to both sides of the equation:\n\nThe x term here is -6x\nThe coefficient is -6\nHalf of -6 is -3 (Remember this number -3)\n-3 sqared is 9. Add 9 to both sides of the equation:\n\nx^2 - 6x + 9 = 16 + 9\n\n#3. If you did #2 correctly, then the left side is a perfect square and the constant term is -3 (remember -3?)\n\n(x - 3)^2 = 25\n\n(If you're not sure, multiply it out and see if you get what you had before.)\n\n#4. Take the square root of both sides. Remember, it could be negative or positive.\n\nx - 3 = 5\nx - 3 = -5\n\n#5. Solve your two equations\n\nx = 8\nx = -2\n\n#6. Put your answers back into the original equation and see if you got the right answer. \n\n\nStep 4. \n\nIf you did Step 3, you already have the answer, and you're already done. If you didn't do Step 3, use the quadratic formula to get the answer. You can find the quadratic formula here:\n\nhttp://en.wikipedia.org/wiki/Quadratic_formula\n\nMemorize the quadratic formula and never forget it.  Write it on your forhead until you have it memorized.
287286
This is a quadratic equation hence there are two factors.\n\nFirst factorise the equation: (3x-1)(2x-3)=0\nThen each factor = 0\nHence x=1/3 (from the first factor)\nand x=3/2 (from the second factor)\n\nyou can recheck this by substituting these two values in the original equation and it shoudl add up to zero.\n\nThe factors are basically points on the curve that cut the x axis (where y=0)
1012771
First are the answers, and their individual in-depth explanations are below.  Hope this helps!\nA. S=5\nB. X=-8\nC. Y=-3X+1\nD. It is a number that cannot be expressed as the ratio of two integers.\nE. A trinomial is a polynomial expression with three terms (after simplification).  The trinomial you are looking for is 3x^2+11x-4.\nF. 2 and (N^2 - 3). Factor out all common factors.\n\n\n\nA. Let K be the constant of variation.  T=KS. 12 =15K. K=4/5.  Now T=(4/5)S.  Substitue the values you've been given.  4=(4/5)S.  S=5\n\nB. X = -8.  An undefined expression includes division by zero, since this is impossible.  To make the denominator zero, X+8=0.\n\nC. Actually, it's Y=-3x+1.  This is just slope-intercept form, y=mx+b, where m=slope and b=y-intercept.  But think about it.  Y-intercept is where the graph crosses the y intercept, so the x-coordinate has to be zero. If you substitute 0 for x in the equation, you get Y=1, so this is correct.\n\nD. It is a number that cannot be expressed as the ratio of two integers.  It is still a real number, though (so no square roots of negative numbers!)  There are two types of irrational numbers: Radicals (the square root of five, the fifth root of 11) and Transcendental numbers (pi, e)\n\nE. A trinomial is a polynomial expression with three terms (after simplification).  For example, x^2 +x+1 or y^21+13y^3+231.  Each term has a different exponent of the variable (Including zero.  That is, y^0 =1)\nTo get the trinomial you are looking for, use the FOIL technique for multiplying binomials.  Multiply the First terms 3x*x (=3x^2). Outside terms are 3x*4 (=12x) and Inside terms are (-1)*x (=-x).  Finally, the Last terms together are (-1)*4 (=-4).  Now add all of these products together.  3x^2+12x+-x-4 = 3x^2+11x-4.\n\nF. Factor out all common factors.  Your original equation is 2n^2-6.  2n^2 = 2*n*n.\n6= 2*3.  \nThe only factor these have in common is two.  Take that out:\n2*(n^2-3)\n\n\n[edit] I just successfully finished an honors Algebra 2 course and I consulted my textbook to answer this question, so I know what I'm talking about!
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^_^
1164880
If x-2 and x+1 is a factor, then P(2) = 0 and P(-1) = 0\nP(2) = 2^4 + a*2^3 - 2^2 +b*2 - 12\n= 16 + 8a - 4 + 2b - 12\n= 8a + 2b = 0\nP(-1) = 1 - a - 1 - b - 12\n= -a - b - 12 = 0\nYou now have a system of equations.\n8a + 2b = 0\n-a - b - 12 = 0\nSolve by substitution. a = 4 and b = -16\nNow use synthetic division or long division to get the other roots.\nx = 2 and x = -1 are given.  The other roots are -3 and -2.\n\n2) Not sure\n\n3) Use sythetic division or long division.\nYou will get that f(x) divided by (x+2) is 2x^2 - 2ax + 3a.\nIf -2 is the only real root, then the other roots must be imaginary.\nIf the roots of a quadratic equation are imaginary, then the discriminant is less than 0.\ndiscriminant = b^2 - 4ac, where a = 2, b = -2a and c = 3a\nb^2 - 4ac = (-2a)^2 - 4(2)(3a)\n= 4a^2 - 24a <0\nFactor:\n4a(a-6) < 0\nIf a<0 or a>6 then 4a(a-6) is positive.\nIf 0<a<6, then 4a(a-6) is negative.\nSo 0<a<6.\n\n4) Not sure, but I'll think about it.
825052
The first step would be to try to factor the equation, so you'd want to get all of the terms on the left side of the equation:\nx^2 + 4x -3 = 0\n\nSince this won't easily factor, you need to use a different method.  I'm sure someone will suggest the quadratic equation, which is fine _if you can remember it_.  I much prefer an alternate method, "completing the square," because it's intuitive enough that you can remember how to work out the process even if you forget the steps.\n\n1. Start by moving the constant to the right side of the equation:\nx^2 + 4x = 3\n\n2. Then divide the coefficient of x (in this case, 4) by two and square it, adding it to both sides of the equation:\nx^2 + 4x + 4 = 7\n\n3. Factor the left side of the equation to (x + a)^2, where 'a' will be the number you just squared:\n(x + 2)^2 = 7\n\n4. Take the square root of both sides, remembering that you'll have both a positive and a negative root:\nx + 2 = +/- (sqrt)7\n\n5. Move the constant on the left to the right, and you have your two solutions:\nx = -2 +/- (sqrt)7
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 )
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)
241036
I'm assuming you're looking for information on factoring polynomials since it's a College Algebra courese.  \nThis website is very helpful, with the steps for factoring laid out, and it also provides websites you can use if you need more information.  Good luck!\n\nhttp://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut7_factor.htm
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.