Question

Find all the zeroes of the polynomial function f(x)=x^3-5x^2+6x-30 in you use synthetic division show all three lines of numbers

Answer 1

Answer:

One real root  $x_{1}=5$

Two imaginary roots $x_{2}=2.449489i$             $x_{3}=-2.449489i$          

Step-by-step explanation:

Given polynomial,

$f\left ( x \right )=x^{3}-5x^{2}+6x-30$

$f\left ( x \right )=x^{3}-5x^{2}+6x-30=0$

$f\left ( x-k \right )=0$

Apply hit and trial method  to find zero

$x=5$

$f\left ( 5 \right )=5^{3}-5\times 5^{2}+6\times 5-30$

       $=125-125+30-30$

$f\left ( 5 \right ) =0$

This polynomial has one factor is     $\left ( x-5 \right )$

We can find another factor

$f\left ( x \right )/\left ( x-5 \right )=\left ( x^{3}-5x^{2}+6x-30\right )/\left ( x-5 \right )$

                     $=x^{2}+6$

$f\left ( x \right )=\left ( x-5 \right )\left ( x^{2}+6 \right )=0$

          $\left ( x-5 \right )\left ( x^{2}+6 \right )=0$

               $\left ( x-5 \right )=0$

Real root           $x_{1}=5$

    $\left ( x^{2}+6 \right )=0$                 $\left ( x^{2}+6 \right )$  is always positive so it have no real roots

Roots are imaginary

$x_{2}=2.449489i$

$x_{3}=-2.449489i$

$f\left ( x \right )$ has one real root and two imaginary roots