what is the easiest way to factor trinomials?
374428
The most strait forward way is to use the quadratic formula. You just do simple adding, powering, rooting, and you get the answer. It works the 100% of the time. There is also the filling-in-the-square, but that's too much work compaired to the quadratic formula.
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)
815517
OK First you take out the common factor which is y. After you have done that.you have (y^2-12y+36). After this, it becomes simple factoring. The answer is\ny(y-6)(y-6)\nor also known as y(y-6)^2
565133
We start by factoring the polynomial.  A little trial and error (testing the factors of 2, which is the constant term) reveals that 2 is a root.  So, we know that (x-2) is a factor.\n\nDividing the polynomial by (x-2) using synthetic division, we find the other factor is 6x^2+x-1.  This can be factored as (3x-1) (2x+1).\n\nSo, the equation, in factored form, is\n(x-2)(3x-1)(2x+1)=0\n\nSo, the roots are x=2, x=1/3, and x=-1/2.
908426
Plug in each of the four answers.  The only one that works is (2), because you get\n\n | 3 (-2/3) - 3 | = 5\n | -2 - 3 | = 5\n | -5 | = 5\n 5 = 5
119973
(x+3)(x^2-x+4)/(x+3) common factor\n\nmultiply by x+3\n\nx^2-x+4
630347
Both of your examples are perfect square trinomials.\n\n(ax)^2 + 2(a)(b)x + b^2 = (ax + b)(ax + b) = (ax + b)^2\n\n9y^2 - 48y + 64 \n= (3y)^2 + 2(3)(-8)y + (-8)^2\n= (3y - 8)^2\n\n16x^2 + 24x + 9 \n= (4x)^2 + 2(4)(3)x + 3^2\n= (4x + 3)^2
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.
916165
Sorry, there isn't one.  If you meant "3x + 9" instead of 3x, then x + 3 is a common factor.
791576
I'm going to assume the placement of parenthesis.\n(x^2 – 3x + 2)/(8x - 8) ÷ (x^2 – 4)/(5x +10)\n\nStep 1: Flip the second fraction, so that you have multiplication instead of division.\n(x^2 – 3x + 2)/(8x - 8) * (5x +10)/(x^2 – 4)\n\nStep 2: Factor.\nx^2 – 3x + 2 = (x - 2)(x - 1)\n8x - 8 = 8(x-1)\n5x +10 = 5(x + 2)\nx^2 – 4 = for a^2 - b^2, the factors are (a + b)(a - b). So: (x + 2)(x -2)\n\n(x - 2)(x - 1) / 8(x-1) * 5(x + 2) / (x + 2)(x -2)\n\nStep 3: Cancel out, multiply\n(x - 2) / 8 * 5 / (x -2)\n5 / 8 (solution!)
816805
i doubt u got it right. this is how u do it:\n\nfirst u group them \n\n(x³-3x²)+(4x-12)\n\nthen simplify each binomial\n\nx²(x-3)+4(x-3)\n\nthen since (x-3) are the same on both of them u group them together\n\n(x²+4)(x-3)\n\nand there u have it
760715
If it is divisible by x - 3, then (x - 3) is a factor of that polynomial, and if x - 3 = 0, x = 3.  \n\nIf we substitute x = 3 to the polynomial, then it must yield 0 (since 3 is a root).\nx³ + 3x² + kx - 24\n\nThus,\n(3)³ + 3(3)² + k(3) - 24 = 0\n\nThen,\n27 + 27 + 3k - 24 = 0\n\nTranspose,\n3k = -30\n\nThus,\nk = -10\n\nThe polynomial then becomes\nx³ + 3x² - 10x - 24 = 0\n\nWe then divide it by x - 3, to factor it out.  The easiest method goes like this\nx³ - 3x² + 6x² -18x + 8x - 24 = 0\n\nNow, we group\n(x³ - 3x²) + (6x² - 18x) + (8x - 24) = 0\n\nWe factor out from each group\nx²(x - 3) + 6x(x - 3) + 8(x - 3) = 0\n\nWe then factor out x - 3\n(x - 3)(x² + 6x + 8) = 0\n\nWe can factor the quadratic trinomial\n(x - 3)(x + 2)(x + 4) = 0\n\nTherefore, your polynomial is also divisible by (x + 2) and (x + 4).\n\n^_^
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 )
697928
First you group the 'like terms'; in your example, the first two and the last two. Then you see the common factor and pull it out\n\ny^2(x^2-1) - 15 (x^2-1)\nNow you see the common term in both and pull that out\n(x^2-1) (y^2-15)\n\nthus you have your two factors of the polynomial
902478
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
954300
ok\nso its factoring.this is hella easy.10x+40 is 10(x+4)\nso the answer is 10(x+4)there you go
775090
You have to find the two polynomials that when multiplied by each other will equal the trinomial.\n\nx2 + 8x - 9\n\n(x + 9) (x - 1)\n\nx*x + -1x + 9x -9\n\nx2 + 8x - 9