Respuesta :

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.

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