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 August 29th, 2010, 07:55 PM #1 Member   Joined: Jul 2010 Posts: 69 Thanks: 0 Solve Cubic and Higher Equations Wow this book is getting really hard, I only have 8 pages left. Hopefully I can get this done tonight. Here we have: 3x^3 - x^2 - 27x=15? August 29th, 2010, 08:13 PM #2 Senior Member   Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs Re: Solve Cubic and Higher Equations There are three real solutions: x ? -2.48151800622656993853 x ? -0.591509729873239929878 x ? 3.40636106943314320174 Are you sure you copied the problem correctly? I would think it would factor (i.e. you wouldn't be expected to use numerical root finding techniques). August 29th, 2010, 08:15 PM #3 Member   Joined: Jul 2010 Posts: 69 Thanks: 0 Re: Solve Cubic and Higher Equations Hey Mark, I forgot to put the 9 in front of the x squared: 3X^3 - 9X^2 - 27X = 15 Additionally, the answers in the book is -1 or 5 so would you or someone be able to help me out please? August 29th, 2010, 08:44 PM #4 Senior Member   Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs Re: Solve Cubic and Higher Equations Ok, that makes more sense! 3X^3 - 9X^2 - 27X - 15 = 0 For this one, factoring by grouping as it is given will not work, you have to look at the middle two terms and break them up so that you then can factor by grouping. First, look at -9x^2. You need to express this term as the sum of two terms that will go with both the first and last terms, while at the same time writing 27x as the sum of two terms that will also go with the first and last terms. -9 = 6 - 15 and -27 = 3 - 30. Notice that 6 is 2·3, -15 is 1·-15, 3 is 1·3 and -30 is 2·-15, so these will work. We can now write the equation as 3x^3 + 6x^2 + 3x - 15x^2 - 30x - 15 = 0 3x(x^2 + 2x + 1) - 15(x^2 + 2x + 1) = 0 (3x - 15)(x^2 + 2x + 1) = 0 3(x - 5)(x + 1)˛ = 0 x - 5 = 0 and x + 1 = 0 x = 5, x = -1, x = -1 -1 is said to be a root of multiplicity two, just in case you were wondering why this cubic seems to only have two roots. August 29th, 2010, 09:32 PM #5 Member   Joined: Jul 2010 Posts: 69 Thanks: 0 Re: Solve Cubic and Higher Equations Hopefully I will be able to figure that out, thank you. August 29th, 2010, 11:06 PM #6 Global Moderator   Joined: Dec 2006 Posts: 21,128 Thanks: 2336 Dividing by 3 and rearranging gives xł - 3x˛ - 9x - 5 = 0. If that equals (x - a)(x - b)(x - c), a + b + c = 3 and abc = 5, so the roots might be 5, -1 and -1. To verify that those are indeed the roots, expand (x + 1)˛(x - 5). Tags cubic, equations, higher, solve ,

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# my math forum factoring cubic equations for this one factoring by grouping as it is given will not work you have to look at the middle two terms

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