Answer:
[tex]x_{1} =5\\x_{2}=i\sqrt{6}\\x_{3}=-i\sqrt{6}\\[/tex]
Step-by-step explanation:
Unlike quadratic equations, cubic equations might be harder to solve. Therefore, one useful approach is to group and factor terms:
[tex]x^3-5x^2+6x-30=0\\x^2(x-5)+6(x-5)=0[/tex]
the term (x-5) now appears on both parts of the equation and can be factored as follows:
[tex](x^2 +6)*(x-5)=0[/tex]
From here, we can find all three roots to the function:
[tex](x-5)=0\\x_{1} =5\\x^2+6=0\\x=\sqrt{-6}[/tex]
The only real root is 5 since there are no real square roots for negative numbers, the complex roots are:
[tex]x=\sqrt{6}\sqrt{-1}\\x_{2}=i\sqrt{6}\\x_{3}=-i\sqrt{6}\\[/tex]
The zeroes of the polynomial funtion are:
[tex]x_{1} =5\\x_{2}=i\sqrt{6}\\x_{3}=-i\sqrt{6}\\[/tex]
