Answer:
[tex](x - 2)[/tex] isn't a factor of [tex]f(x) = x^{3} - 2\, x^{2} + 2\, x + 3[/tex].
Step-by-step explanation:
By the factor theorem, for any constant [tex]c[/tex], [tex](x - c)[/tex] is a factor of polynomial [tex]f(x)[/tex] if and only if [tex]f(c) = 0[/tex]. Note that [tex]f(c) = 0\![/tex] means that substituting all [tex]x[/tex] in [tex]f(x)\![/tex] with [tex]c[/tex] and evaluating gives [tex]0[/tex].
For example, the polynomial in this question is [tex]f(x) = x^{3} - 2\, x^{2} + 2\, x + 3[/tex]. The question is asking whether [tex](x - 2)[/tex] is a factor of [tex]f(x)[/tex].
By the factor theorem with [tex]c = 2[/tex], [tex](x - 2)\![/tex] would indeed be a factor of [tex]f(x)\![/tex] if and only if [tex]f(2) = 0[/tex]. To find the value [tex]f(2)\![/tex], simplify replace all "[tex]x[/tex]" in the definition of [tex]f(x)\![/tex] with [tex]2[/tex]:
[tex]\begin{aligned}f(2) &= 2^{3} - 2\times (2^{2}) + 2\times 2 + 3 \\ &= 2^{3} - 2^{3} + 7 \\ &= 7\end{aligned}[/tex].
In other words, [tex]f(2) \ne 0[/tex]. By the contrapositive of factor theorem, [tex](x - 2)[/tex] would not be a factor of [tex]f(x)[/tex].
