obrozenecy6

2021-12-31

Using the Fundamental Theorem of Algebra, complete the following exercise. Show your work.
Determine how many, what type, and find the roots for $f\left(x\right)={x}^{3}-5{x}^{2}-25x+125$.

temnimam2

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

Karen Robbins

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

karton

Given that the function
$f\left(x\right)={x}^{3}-5{x}^{2}-25x+125$
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
$f\left(x\right)={x}^{2}\left(x-5\right)-25\left(x-5\right)$
$=\left({x}^{2}-25\right)\left(x-5\right)$
$=\left(x+5\right)\left(x-5{\right)}^{2}$
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

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