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How do you find the number of complex, real and rational roots of #x^3-6x^2-7x-12=0#?

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Feb 1, 2018

Answer:

This cubic has one positive irrational real root and two non-real complex roots.

Explanation:

Given:

#f(x) = x^3-6x^2-7x-12=0#

Rational root theorem

By the rational root theorem, any rational root of this polynomial is expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-12# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational roots are:

#+-1, +-2, +-3, +-4, +-6, +-12#

Descartes' Rule of Signs

The pattern of signs of the coefficients of the polynomial is #+ - - -#. With one change of signs Descartes' Rule of Signs tells us that this cubic has exactly one positive root.

We find that:

#f(6) = -54 < 0#

#f(12) = 768 > 0#

So the real positive root cannot be rational and must be an irrational number in #(6, 12)#.

Reversing the signs on the terms of odd degree, we get the pattern #- - + -#. With two changes of sign, we can deduce that this cubic has #0# or #2# negative real roots.

Discriminant

The discriminant #Delta# of a cubic polynomial in the form #ax^3+bx^2+cx+d# is given by the formula:

#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#

In our example, #a=1#, #b=-6#, #c=-7# and #d=-12#, so we find:

#Delta = 1764+1372-10368-3888-9072 = -20192#

Since #Delta < 0# this cubic has #1# Real zero and #2# non-Real Complex zeros, which are Complex conjugates of one another.

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