Answer: x = 2
Step-by-step explanation:
Given polynomial function,
[tex]f(x) = 3x^3-8x^2 + 3x + 2[/tex],
According to question, If x = 1 is a zero of the polynomial function [tex]f(x) = 3x^3-8x^2 + 3x + 2[/tex]
Therefore, (x - 1) is the factor of [tex]f(x) = 3x^3 - 8x^2 + 3x + 2[/tex]
By dividing [tex]3x^3- 8x^2 + 3x + 2[/tex] by (x - 1) (mentioned in below diagram)
We get, [tex]3x^2-5x-2[/tex].
Therefore, [tex]3x^3 - 8x^2 + 3x + 2=(x - 1)(3x^2-5x-2)[/tex]
⇒[tex]3x^3 - 8x^2 + 3x + 2=(x - 1)(3x^2-6x+x-2)[/tex] (by grouping method)
⇒[tex]3x^3- 8x^2 + 3x + 2=(x - 1)(3x+1)(x-2)[/tex]
Thus, x-2 is a factor of [tex]3x^3- 8x^2 + 3x + 2[/tex],
Therefore, x=2 is a zero of [tex]3x^3- 8x^2 + 3x + 2[/tex]
