What are all the possible rational zeros for #f(x)=2x^4-6x^2+5x-15# and how do you find all zeros?

1 Answer
Aug 11, 2017

Answer:

"Possible" rational zeros are:

#+-1/2, +-1, +-3/2, +-5/2, +-3, +-5, +-15/2, +-15#

For actual zeros, read on...

Explanation:

Given:

#f(x) = 2x^4-6x^2+5x-15#

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 #-15# 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, +-5/2, +-3, +-5, +-15/2, +-15#

With the function as specified, none of these is a zero.

If we graph #f(x)# then it looks like this:

graph{2x^4-6x^2+5x-15 [-5, 5, -35, 25]}

There are two irrational zeros, one somewhere in the interval #(-5/2, -2)#, the other in the interval #(3/2, 2)#.

Typically for a quartic, these are possible but somewhat messy to find algebraically. If you would really like to see a solution, please request in the comments.

I suspect a typo in the question.

Consider the function:

#f_1(x) = 2x^3-6x^2+5x-15#

The rational roots theorem suggests the same possible rational roots, but this time the graph looks like this:

graph{2x^3-6x^2+5x-15 [-5, 5, -35, 25]}

There is one real rational root at #x=3#:

#f_1(3) = 2(color(blue)(3))^3-6(color(blue)(3))^2+5(color(blue)(3))-15#

#color(white)(f_1(3)) = 54-54+15-15#

#color(white)(f_1(3)) = 0#

We can also factor #f_1(x)# by grouping:

#2x^3-6x^2+5x-15 = (2x^3-6x^2)+(5x-15)#

#color(white)(2x^3-6x^2+5x-15) = 2x^2(x-3)+5(x-3)#

#color(white)(2x^3-6x^2+5x-15) = (2x^2+5)(x-3)#

From this we can see the real zero #x=3# and find the complex zeros:

#x = +-sqrt(5/2)i = +-sqrt(10/4)i = +-sqrt(10)/2i#