ANSWER :
-2, 1 and 2
EXPLANATION :
From the problem, we have the function :
[tex]f(x)=x^5-3x^4+2x^3+8x^2-24x+16[/tex]
We can regroup the function :
[tex]f(x)=(x^5-3x^4+2x^3)+(8x^2-24x+16)[/tex]
Factor out x^3 from the first group and 8 from the second group :
[tex]f(x)=x^3(x^2-3x+2)+8(x^2-3x+2)[/tex]
Then rewrite as a product of two polynomials :
[tex]f(x)=(x^3+8)(x^2-3x+2)[/tex]
The first factor is the "sum of two cubes" which can be factored in the form :
[tex]x^3+y^3=(x+y)(x^2-xy+y^2)[/tex]
So that will be :
[tex]\begin{gathered} f(x)=(x^3+8)(x^2-3x+2) \\ f(x)=(x+2)(x^2-2x+4)(x^2-3x+2) \end{gathered}[/tex]
Factor the last parenthesis :
[tex]f(x)=(x+2)(x^2-2x+4)(x-2)(x-1)[/tex]
The zeros of the function is when f(x) = 0
So equate all factors to 0.
x + 2 = 0
x = -2
x - 2 = 0
x = 2
x - 1 = 0
x = 1
x^2 - 2x + 4 = 0
Use the determinant formula to check if it has rational or irrational roots.
[tex]\begin{gathered} b^2-4ac=(-2)^2-4(1)(4) \\ =-12 \end{gathered}[/tex]
Since the result is a negative number, it has imaginary roots.
So the real and rational roots are -2, 1, and 2
