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solve cubic with no rational zeros |
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| Sep18-12, 11:19 PM | #1 |
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solve cubic with no rational zeros
1. The problem statement, all variables and given/known data
x^3-8x+10=0 2. Relevant equations 3. The attempt at a solution By dividing by x-1 (1 is a factor of the solution), I got (x-1)(x2+x-7)+3+10=0 which equals x^3-8x+20=0 I think the solution would be imaginary but I used a graphing calculator and there is one solution but I don't know how to calculate it. |
| Sep19-12, 12:44 AM | #2 |
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Mentor
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| Sep20-12, 02:08 PM | #3 |
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-(1/3)*(135+3*sqrt(489))^(1/3) - 8/(135+3*sqrt(489))^(1/3) and a decimal approximation to that is -3.318628218. |
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