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2007-05-28 10:59:33 · 4 answers · asked by dACTIVEdoll 1 in Science & Mathematics Mathematics

4 answers

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 · answer #1 · answered by Orinoco 7 · 0 0

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 · answer #2 · answered by Pascal 7 · 0 0

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 · answer #3 · answered by Puggy 7 · 0 0

f(x)= x^3+x^2+x-12
=

2007-05-28 18:09:37 · answer #4 · answered by Anonymous · 0 0

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