Using the Fundamental Theorem of Algebra, complete the following exercise. Show your work. Determine how...

Step 1
Fundamental Theorem of Algebra, states if f(x) is a polynomial of degree n, where ${\displaystyle n>0}$, then f has at least one zero in the complex number system.
The polynomial given to us is ${\displaystyle f\left(x\right)=x^3-5x^2-25x+125}$.
Step 2
the highest power of this polynomial is 3 so, there will be 3 roots to this polynomial.
now, by factorizing the polynomial we get
${\displaystyle f\left(x\right)=(x-5)(x^2-25)}$
${\displaystyle f\left(x\right)=(x-5)(x-5)(x+5)}$
${\displaystyle f\left(x\right)={(x-5)}^2(x+5)}$
therefore, we can see that the roots of the polynomial are real and the roots of the polynomial are ${\displaystyle x=5,-5}$
and the multiplicity of the roots ${\displaystyle x=-5is2}$

by grouping method:
${\displaystyle f\left(x\right)=x^3-5x^2-25x+125}$.
${\displaystyle =x^2(x-5)-25(x-5)}$
${\displaystyle f\left(x\right)=(x-5)(x^2-25)=(x-5)(x-5)(x+5)}$
${\displaystyle f\left(x\right)={(x-5)}^2(x+5)}$

Given that the function
$f(x)=x^3-5x^2-25x+125$
We have to find the roots of f(x)
Let us try to factorize the function to find the roots
We can group two by two and find out
$f(x)=x^2(x-5)-25(x-5)$
$=(x^2-25)(x-5)$
$=(x+5)(x-5{)}^2$
We find that the roots are -5,5,5
Or -5 with a multiplicity of 1 and 5 with a multiplicity of 2 are the roots of the equation