Need help factoring x^3-6x^2-3x+18=0?
825419
x^3 - 6x^2 - 3x + 18 \n= x^2 (x -6) - 3(x-6) \n= (x^2 - 3)(x - 6)
843398
6x(x-3y)-3(6x^2-xy) = 6x^2-18xy-18x^2+3xy=-12x^2-15xy ans.
973855
3x² + 18x = 0 --- First, I would take 3x out of the expression on the left...\n(3x)(x +6) = 0 --- Now, the left side is factored.\nTo solve, we know that "3x" and/or "x + 6" MUST equal 0.\n3x = 0 --- Divide both sides by 3...\nx = 0 --- One of the answers...\n\nx + 6 = 0 --- Subtract 6 from both sides...\nx = -6 --- The other answer...\n\nAnswer: x = 0 or x = -6
526107
You are only allowed to factor if the equation is equal to zero; your's is equal to 4.  You need to solve "x^3 - 6x^2 +9x - 4 = 0"  using synthetic division.
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^_^
785437
3x^2-75 = 0\n3x^2 = 75\nx^2 = 25\nx = +/-5\n\nThese ones are fun since there are two solutions that you can easily find by taking square roots.  The factored form is:\n\n3(x-5)(x+5)\n\nHope this helps.
204966
I can solve it if you'll clarify whether the equation is\n2x^2 = 17/3x + 1 OR 2x^2 = (17/3)x + 1 OR 2x^2 = 17/(3x+1).\nI'm not posting a bunch of steps for three different equations.\n\nI'll post this one (it's the easiest to explain):\n\n2x^2 = (17/3)x + 1\n\nMove all the terms to left side of the equal sign.\n2x^2 - (17/3)x - 1 = 0\n\nClear the fraction by multiplying all terms by 3.\n3(2x^2) - 3(17/3)x - 3(1) = 3(0)\n\nThe equation is now:\n6x^2 - 17x - 3 = 0\n\nNow you multiply the coefficient of the first term by the third term:\n(6)(-3) = -18\n\nNow you need two factors of -18 that when added equal -17.\nIn this case, they are -18 & 1.\nSubstitute them for the middle term.\n6x^2 + (-18 + 1)x - 3 = 0\n\nRemove the parentheses and distribute the x. The equation becomes:\n6x^2 - 18x + x - 3 = 0\n\nExtract the greatest common factor for the first two terms:\n6x(x - 3) + x - 3 = 0\n\nNow extract the greatest common factor for the last two terms:\n6x(x - 3) + 1(x - 3) = 0\n\nRewrite the equation as its factors:\n(6x + 1)(x - 3) = 0\n\nSet the value of each factor to 0:\n6x + 1 = 0 & x - 3 = 0\n\nSolve each:\n\n6x + 1 = 0\n6x = -1\n6x/6 = -1/6\nx = -1/6\n\nx - 3 = 0\nx = 3\n\nSo, x = -1/6 & x = 3
735202
You mean find solutions for x when 2x^2+12x+18=0.\n\n  First we can divide thru by 2 just to make it simpler.\n x^2+6x+9=0 , then factor and get\n(x+3)^2=0\nso x=3.
379228
Factor it by "FOIL":\n\n(x^1/3 - 2)(x^1/3 - 3) = 0\n\nThen x^1/3 = 2 or x^1/3 = 3\n\nDo both sides of each equation to the third power :\n\nx = 2^3 or x = 3^3\n\nSo x = 8 or 27.
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
1079532
2xy^2(3y^2)^(1/3)\n\ndivide the exponents by three, leave the remainder in the root.\nfind the lowest cubic factor in 24 (8=2^3) and take it out, leave the 3 in.
1098481
6(2x + 8) - 6 =0\n12x + 48 - 6 = 0\n12x + 42 = 0\n12x = -42\nx = -42/12\nx = -3.5
714160
x^2+6x+8/x^2-4-21 / x^2-4/x^2-7x\n=x^2 + 6x +(8/x^2) - 4 - (21/x^2) - (4/x^2) - 7x\n=x^2 + 6x -4 -7x +(8/x^2) - (21/x^2) - (4/x^2) \n=x^2 + 6x -4 -7x +[8-21-4]/x^2\n=x^2 + 6x -4 -7x -17/x^2\n=(x^2)*[x^2 - x -4  -17]/x^2\n=(x^2)*(x^2) - (x^2)*(x) -4*(x^2)  -17]/x^2\n=(x^4) - (x^3) - 4(x^2)x -17]/x^2
252087
(12x^3-18x^2)-8x+12=\n\n6x^2(2x-3)-4(2x-3)=\n\n(2x-3)(6x^2-4)=\n\n(2x-3)*2*(3x^2-2)=\n\n2(2x-3)(3x^2-2)
10610
X^2-9X+18\n\nShort and simple just factor it FOIL
890098
I enjoy algebra, and this is easily my favorite type of problem.  So I hope you, too, one day enjoy them for their inner beauty.\n\nFor now, let's tackle this "factoring problem."  We may wish to use the method of "synthetic division" which is arguably easier than "long division of polynomials."\n\nThink back to grade school where you learned that 6 = 2 * 3.  This simple statement enabled you to 'deconstruct' the number 6 as either "two times three" or "three times two".  Even if you forgot one of the 'factors' you could conjure it up again quickly by asking yourself, "What is six divided by two?" for example.\n\nIn problem (a), you are essentially asking, "What are the coefficients of P(x) if the only factor we know is (x-1)?"\n\nLet's set up synthetic division (see link below), by placing all the coefficients of P(x), and using 1 (the zero implied by (x-1)) as the divisor:\n\n1 ||  a  b  c  d\n___________\n.....a\n\n1 ||  a  b  c  d\n_____a_____\n.....a (a+b)\n\n1 ||  a  b        c       d\n_____a___(a+b)____\n.....a (a+b) (a+b+c)\n\n1 ||  a  b        c                d\n_____a___(a+b)___(a+b+c)_\n.....a (a+b) (a+b+c) (a+b+c+d)\n\nNote that this last term must equal 0 since it is assumed that (x-1) is a factor ... thus, the "remainder" term must equal zero!\n\nThere's the result you were after:  a+b+c+d = 0\n\nHope this helps you tackle the others!
919554
Factor out 3.\n3(x^2+2x+1)=0\n3(x+1)^2=0\nThe only answer then is x=-1 to make the expression equal 0.