There's a single real root somewhere near x = 1.8761485603331087
and 2 imaginary roots approx
x = -1.4380742801665543 +/- i* 2.080390431024078
2007-05-28 11:04:34
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answer #1
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answered by Orinoco 7
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Per the rational root theorem, if this function has any rational roots, they must be of the form p/q, where p and q are integers, p divides the constant coefficient (which is -12), and q divides the leading coefficient (which is 1). The possible rational roots are therefore:
±1, ±2, ±3, ±4, ±6, and ±12
2007-05-28 18:18:47
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answer #2
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answered by Pascal 7
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f(x) = x^3 + x^2 + x - 12
For this cubic ax^3 + bx^2 + cx + d, list all possible factor of d over all possible factors of a. In this case, a = 1, so we will be dealing with all integers.
The factors of -12 are 1, -1, 2, -2, 3, -3, 4, -4, 6, -6, 12, -12
Those are the possible rational zeros.
2007-05-28 18:16:11
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answer #3
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answered by Puggy 7
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f(x)= x^3+x^2+x-12
=
2007-05-28 18:09:37
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answer #4
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answered by Anonymous
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