91Ó°ÊÓ

The given function is$f\left(x\right)=x^3-4x^2+25x-100$

The given function has a degree 3. So, there will be a maximum of 3 zeroes.

We will use the Descartes rule of signs to find the number of positive and negative real zeroes of the function.

The function has three sign variation. So, there is 3 or 1 positive real zero.

$$\begin{gathered} f(-x)={(-x)}^3-4{(-x)}^2+25(-x)-100 \\ =-x^3-4x^2-25x-100 \end{gathered}$$

Since, there is no sign change in $f(-x)$, there is no negative real zero.

We will use the rational zeroes theorem to find the potential rational zeroes of the function.

List the integers which are factors of constant term $-100$

$p:Â\pm 1,Â\pm 2,Â\pm 4,Â\pm 5,Â\pm 10,Â\pm 20,Â\pm 25,Â\pm 100$

List of integers which are factors of leading coefficient 1.

$q:Â\pm 1$

List the possible potential rational zeroes which is the ratio $\frac{p}{q}$

$\frac{p}{q}:Â\pm 1,Â\pm 2,Â\pm 4,Â\pm 5,Â\pm 10,Â\pm 20,Â\pm 25,Â\pm 100$

Thus, the potential rational zero of function are 1,2,4,5,10, 20,25,100.

Check whether 4 is a zero of function or not.

Thus, the depressed equation is $x^2+25=0$

Find the remaining zeroes by solving the equation,

$$\begin{gathered} x^2+25-25=0-25 \\ {x}^2=-25 \\ x=Â\pm 5i \end{gathered}$$

The three complex zeroes of the function are$\left\{4,5i,-5i\right\}$and the factored form is$f\left(x\right)=(x-4)(x+5i)(x-5i)$