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djyliett [7]
3 years ago
13

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

Mathematics
1 answer:
Sloan [31]3 years ago
6 0

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

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