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

##### 1 Answer

#### Answer:

This quintic has no rational zeros. It has one irrational Real zero and

#### Explanation:

By the rational root theorem, any *rational* zeros of

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

Looking at the derivative of

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

So

#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

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

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