What are all the possible rational zeros for #f(x)=4x^5+4x^2-9# and how do you find all zeros?

1 Answer
Dec 8, 2016

Answer:

This quintic has no rational zeros. It has one irrational Real zero and #4# non-Real complex ones.

Explanation:

#f(x) = 4x^5+4x^2-9#

By the rational root 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 #-9# and #q# a divisor of the coefficient #4# of the leading therm.

That means that the only possible rational zeros are:

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

We find:

#f(1) = 4+4-9 = -1 < 0#

#f(3/2) = 4(243/32)+4(9/4)-9 = 243/8 > 0#

So #f(x)# has an irrational zero in #(1, 3/2)#

Looking at the derivative of #f(x)# we find:

#f'(x) = 20x^4+8x = 4x(5x^3+2)#

So #f(x)# has turning points at #x=0# and #x=-root(3)(2/5)#

#f(0) = -9 < 0#

#f(-root(3)(2/5)) = 4(root(3)(2/5))^2(2/5+1)-9 = 28/5(root(3)(2/5))^2-9 < 6-9 < 0#

So the only Real zero of #f(x)# is the irrational one in #(1, 3/2)#

Typically for a quintic, it is not possible to express the zeros in terms of radicals or elementary functions (including trigonometric, logarithmic or exponential functions).

The other #4# Complex zeros will occur as two pairs of complex conjugates.

It is possible to find numerical approximations using algorithms such as Durand-Kerner.