How do you find the number of complex, real and rational roots of #4x^3-12x+9=0#?

1 Answer
Dec 3, 2016

Answer:

This cubic has one negative irrational Real root and two Complex ones.

Explanation:

#f(x) = 4x^3-12x+9#

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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=4#, #b=0#, #c=-12# and #d=9#, so we find:

#Delta = 0+27648+0-34992+0 = -7344#

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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Descartes' Rule of Signs

The signs of the coefficients of #f(x)# are in the pattern #+ - +#. With #2# changes of sign, #f(x)# has #0# or #2# positive Real zeros. Since we know that it does not have #2#, it must have #0# positive Real zeros. The signs of the coefficients of #f(-x)# are in the pattern #- + +#. With one change of sign, that means that #f(x)# has exactly one negative Real zero.

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Rational root theorem

Since all of the coefficients of #f(x)# are integers and the powers of #x# are in standard descending order, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #9# and #q# a divisor of the coefficient #4# of the leading term.

That means that the only possible rational zeros are:

#+-1/4, +-1/2, +-3/4, +-1, +-3/2, +-3, +-9/2, +-9#

Note that we already know that the only Real zero is negative, so if there is a rational zero then it is one of:

#-1/4, -1/2, -3/4, -1, -3/2, -3, -9/2, -9#

We find:

#f(-3) = -63 < 0#

#f(-3/2) = 27/2 > 0#

So the Real zero is somewhere between the possible rational zeros #-3# and #-3/2# and must be irrational.

Conclusion

#f(x)# has one Real negative irrational zero and two Complex zeros.