[tex]f(x) = 0 \text{ when } x \in \{ 1, -3, 10 \}[/tex]
Given:
[tex]f(x) = x^3 -8x^2 -23x +30[/tex]
Note that if [tex]x +3[/tex] is a factor of [tex]f(x)[/tex], then [tex](x +3)a = f(x)[/tex] where [tex]a[/tex] is also a factor of [tex]f(x)[/tex]. This is the same logic as [tex]\blue 3[/tex] is a factor of [tex]12[/tex] so [tex]\blue 3 \times a = 12[/tex] and [tex]a[/tex] would be [tex]4[/tex] which we got it by dividing [tex]12[/tex] by [tex]\blue 3[/tex]. In this case, we can divide [tex]f(x)[/tex] by [tex]x +3[/tex] to get the other factor of [tex]f(x)[/tex] (The solution for that is found on the attached image).
[tex]a = x^2 -11x +10[/tex]
We also have to note that multiplying any number with [tex]0[/tex] gives [tex]0[/tex]. If the value of [tex]x[/tex] makes [tex]x +3 [/tex] equal [tex]0[/tex], [tex](x +3)a[/tex] will be [tex]0[/tex] no matter what the value of [tex]a[/tex] will be. We can see that [tex]x = -3[/tex] makes [tex]x +3 = 0[/tex]. [tex]x = -3[/tex] is one of our zeros.
Now we have to solve at what value of [tex]x[/tex] make [tex]a = 0[/tex].
Solving for [tex]x[/tex]:
[tex]a = 0 \\ x^2 -11x +10 = 0 \\ x^2 -11x = -10 \\ x^2 -11x +\frac{121}{4} = -10 +\frac{121}{4} \\ (x -\frac{11}{2})^2 = -\frac{40}{4} +\frac{121}{4} \\ (x -\frac{11}{2})^2 = \frac{81}{4} \\ x -\frac{11}{2} = \pm \sqrt{\frac{81}{4}} \\ x -\frac{11}{2} = \pm \frac{9}{2}[/tex]
Solving from the Positive Root:
[tex]x -\frac{11}{2} = \frac{9}{2} \\ x = \frac{9}{2} +\frac{11}{2} \\ x = \frac{20}{2} \\ x = 10[/tex]
Solving from the Negative Root:
[tex]x -\frac{11}{2} = - \frac{9}{2} \\ x = -\frac{9}{2} +\frac{11}{2} \\ x = \frac{2}{2} \\ x = 1[/tex]
