How do you find the number of complex, real and rational roots of #2x^4+x^2-x+6=0#?

1 Answer
George C.
Oct 17, 2016

This quartic equation has exactly #4# non-Real Complex roots.

Explanation:

#f(x) = 2x^4+x^2-x+6#

By the rational roots theorem, 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 #6# and #q# a divisor of the coefficient #2# of the leading term.

That means that the only possible rational zeros are:

#+-1/2, +-1, +-3/2, +-2, +-3, +-6#

We could try these, but let's do some more analysis first.

Note that the pattern of signs of the coefficients is: #+ + - +#

By Descartes' Rule of Signs, since there are two changes of sign, that means that #f(x)# has #0# or #2# positive Real zeros.

The pattern of signs of coefficients of #f(-x)# is #+ + + +#. With no changes of sign, that means that #f(x)# has no negative Real zeros.

So the only possible remaining rational zeros are:

#1/2, 1, 3/2, 2, 3, 6#

We could try these, but actually there's a shortcut...

#2x^4+x^2-x+6 = 2x^4+(x-1/2)^2+23/4 >= 23/4#

for all Real values of #x#.

So all of the zeros of #f(x)# are Complex.

By the Fundamental Theorem of Algebra, #f(x)# has exactly #4# zeros since it is of degree #4#.

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