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
"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
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
graph{2x^4-6x^2+5x-15 [-5, 5, -35, 25]}
There are two irrational zeros, one somewhere in the interval
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
#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
#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 = +-sqrt(5/2)i = +-sqrt(10/4)i = +-sqrt(10)/2i#