Answer:
For [tex]x \in\mathbb{C}[/tex]
[tex]x = -2i[/tex]
[tex]x = 2i[/tex]
[tex]x = 7[/tex]
Step-by-step explanation:
[tex]f(x)=x^3-7x^2+4x-28[/tex]
For [tex]f(x) = 0[/tex]
[tex]x^3-7x^2+4x-28 = 0[/tex]
We can easily factor the expression [tex]x^3-7x^2+4x-28[/tex] by grouping.
[tex]x^3-7x^2+4x-28 = (x^3-7x^2)+(4x-28) = x^2(x-7)+4(x-7) = \boxed{(x^2+4)(x-7) }[/tex]
Now we have
[tex](x^2+4)(x-7) = 0[/tex]
Therefore,
[tex]x^2+4 = 0 \text{ or } x-7 = 0[/tex]
So,
[tex]x^2 = -4 \implies x = \pm\sqrt{-4}[/tex]
Once [tex]\sqrt{-1} = i[/tex]
[tex]x = \pm\sqrt{-4} = \pm\sqrt{4}\sqrt{-1} = \pm 2i[/tex]
