Answer:
[tex](x + 2)[/tex] isn't a factor of the polynomial [tex]p(x) = x^{3} + 2\, x^{2} - x - 2[/tex].
Step-by-step explanation:
By the Factor Theorem, for any constant [tex]a[/tex], [tex](x - a)[/tex] would be a factor of a polynomial [tex]p(x)[/tex] if and only if [tex]x = a[/tex] is a root (a zero) of that polynomial ([tex]p(a) = 0[/tex].)
In other words, [tex](x - a)[/tex] is a factor of polynomial [tex]p(x)[/tex] if and only if replacing all mentions of [tex]x[/tex] in [tex]p(x)\![/tex] with the constant [tex]a[/tex] would yield [tex]0[/tex].
For example, in this question, [tex](x + 2)[/tex] could be rewritten as [tex](x - (-2))[/tex] (value of the constant is [tex]a = (-2)[/tex].)
By the Factor Theorem, [tex](x - (-2))[/tex] would be a factor of the polynomial [tex]p(x) = x^{3} + 2\, x^{2} - x - 2[/tex] if and only if replacing all mentions of [tex]x[/tex] in [tex]x^{3} + 2\, x^{2} - x - 2[/tex] with [tex](-2)[/tex] evaluates to [tex]0[/tex].
Replacing all mentions of [tex]x[/tex] in [tex]x^{3} + 2\, x^{2} - x - 2[/tex] with [tex](-2)[/tex] gives:
[tex](-2)^{3} + 2\, (-2)^{2} - (-2) - 2[/tex].
Simplify:
[tex]\begin{aligned}& (-2)^{3} + 2\, (-2)^{2} - (-2) - 2 \\ =\; & (-8) + 8 + 2 - 2 \\ =\; & 0\end{aligned}[/tex].
Since [tex]p(-2) = 0[/tex], [tex](x - (-2))[/tex] (or equivalently, [tex](x + 2)[/tex]) is indeed a factor of the polynomial [tex]p(x)[/tex].
