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August 29th, 2010, 07:55 PM   #1
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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?
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August 29th, 2010, 08:13 PM   #2
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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).
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August 29th, 2010, 08:15 PM   #3
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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?
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August 29th, 2010, 08:44 PM   #4
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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 23, -15 is 1-15, 3 is 13 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.
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August 29th, 2010, 09:32 PM   #5
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Re: Solve Cubic and Higher Equations

Hopefully I will be able to figure that out, thank you.
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August 29th, 2010, 11:06 PM   #6
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Dividing by 3 and rearranging gives x - 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).
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