Answer:
We conclude that option A is true as x = 1 is the root of the polynomial.
Step-by-step explanation:
Given the polynomial
[tex]f\left(x\right)\:=\:x^2\:+\:2x^2\:-\:x-2[/tex]
Let us determine the root of the polynomial shown below.
[tex]\:0=\:x^2\:+\:2x^2\:-\:x-2[/tex]
[tex]0=3x^2-x-2[/tex]
switch sides
[tex]3x^2-x-2=0[/tex]
as
[tex]3x^2-x-2=\left(3x+2\right)\left(x-1\right)[/tex]
so the equation becomes
[tex]\left(3x+2\right)\left(x-1\right)=0[/tex]
Using the zero factor principle
[tex]3x+2=0\quad \mathrm{or}\quad \:x-1=0[/tex]
solving
[tex]3x+2=0[/tex]
[tex]3x=-2[/tex]
[tex]\frac{3x}{3}=\frac{-2}{3}[/tex]
[tex]x=-\frac{2}{3}[/tex]
and
[tex]x-1=0[/tex]
[tex]x=1[/tex]
The possible roots of the polynomial will be:
[tex]x=-\frac{2}{3},\:x=1[/tex]
Therefore, from the mentioned options, we conclude that option A is true as x = 1 is the root of the polynomial.