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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)² = 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 x³  3x²  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)²(x  5). 

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okay that makes more sense 3x^39x^227x15=0,solve cubic equation by grouping,solving cubic equation by grouping,3x^39x^227x15=0,3x^39x^227x15=0 for this one,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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