Answer:
[tex]b = 2[/tex].
Step-by-step explanation:
By the factor theorem, if a monomial of the form [tex](x - a)[/tex] is a factor of the polynomial [tex](3\, x^{3} + b\, x^{2} - 3\, x - 2)[/tex], then substituting in [tex]x = a[/tex] would set this polynomial to [tex]0[/tex].
Notice that in the factor theorem, the coefficient of [tex]x[/tex] in [tex](x - a)[/tex] is [tex]1[/tex], and the sign in front of the constant [tex]a[/tex] is a minus sign. Rewrite the given factor [tex](3\, x + 2)[/tex] in this form to find the value of [tex]a\![/tex]:
[tex]\begin{aligned}& (3\, x + 2) \\ =\; & 3\, (x + (2/3)) \\ =\; & 3\, (x - \underbrace{(-2/3)}_{a}) \end{aligned}[/tex].
Thus, [tex]x = (-2/3)[/tex] should set the polynomial [tex](3\, x^{3} + b\, x^{2} - 3\, x - 2)[/tex] to [tex]0[/tex]:
[tex]\begin{aligned} & (3\, x^{3} + b\, x^{2} - 3\, x - 2) \\ =\; & 3\times (-2/3)^{3} + b\times (-2/3)^{2} - 3 \times (-2/3) - 2 \\ =\; & -3\times (2/3)^{3} + b\times (2/3)^{2} + 3 \times (2/3) - 2 \\ =\; & -\frac{8}{9} + \frac{4\, b}{9} + 2 - 2 \\ =\; & \frac{-8 + 4\, b}{9}\end{aligned}[/tex].
Set this expression to [tex]0[/tex] and solve for [tex]b[/tex]:
[tex]\displaystyle \frac{-8 + 4\, b}{9} = 0[/tex].
[tex]b = 2[/tex].
