Respuesta :
Answer:
One real root [tex]x_{1}=5[/tex]
Two imaginary roots [tex]x_{2}=2.449489i[/tex] [tex]x_{3}=-2.449489i[/tex]
Step-by-step explanation:
Given polynomial,
[tex]f\left ( x \right )=x^{3}-5x^{2}+6x-30[/tex]
[tex]f\left ( x \right )=x^{3}-5x^{2}+6x-30=0[/tex]
[tex]f\left ( x-k \right )=0[/tex]
Apply hit and trial method to find zero
[tex]x=5[/tex]
[tex]f\left ( 5 \right )=5^{3}-5\times 5^{2}+6\times 5-30[/tex]
[tex]=125-125+30-30[/tex]
[tex]f\left ( 5 \right ) =0[/tex]
This polynomial has one factor is [tex]\left ( x-5 \right )[/tex]
We can find another factor
[tex]f\left ( x \right )/\left ( x-5 \right )=\left ( x^{3}-5x^{2}+6x-30\right )/\left ( x-5 \right )[/tex]
[tex]=x^{2}+6[/tex]
[tex]f\left ( x \right )=\left ( x-5 \right )\left ( x^{2}+6 \right )=0[/tex]
[tex]\left ( x-5 \right )\left ( x^{2}+6 \right )=0[/tex]
[tex]\left ( x-5 \right )=0[/tex]
Real root [tex]x_{1}=5[/tex]
[tex]\left ( x^{2}+6 \right )=0[/tex] [tex]\left ( x^{2}+6 \right )[/tex] is always positive so it have no real roots
Roots are imaginary
[tex]x_{2}=2.449489i[/tex]
[tex]x_{3}=-2.449489i[/tex]
[tex]f\left ( x \right )[/tex] has one real root and two imaginary roots
