What is a cubic polynomial function in standard form with zeros 3,-4, and 5?

Hint: To obtain the standard form of the cubic polynomial with zeros 3, -4 and 5 we will use the factored form of the zeros. Firstly as it is given that the zero of the function is 3, -4 and 5 that mean these all values satisfy the function and also the factor form of these values when multiplied give the function. So we will multiply the three factors and put them equal to y and simplify it to get the desired answer.

Complete step by step solution:
The three zeros of the cubic polynomial function are given as below:
$3,-4,5$
All of the three can be written in factored form as below:
$\left( x-3 \right)\left( x+4 \right)\left( x-5 \right)$
Now as we know product of all factor is equal to the function which we let as $y$ we get,
$y=\left( x-3 \right)\left( x+4 \right)\left( x-5 \right)$
On simplifying above value we get,
$\begin{align}
  & y=\left( x-3 \right)\left( \left( x\times x \right)+\left( x\times -5 \right)+\left( 4\times x \right)+\left( 4\times -5 \right) \right) \\
 & \Rightarrow y=\left( x-3 \right)\left( {{x}^{2}}-5x+4x-20 \right) \\
 & \therefore y=\left( x-3 \right)\left( {{x}^{2}}-x-20 \right) \\
\end{align}$
Simplify the third bracket as follows:
$\begin{align}
  & y=x\left( {{x}^{2}}-x-20 \right)-3\left( {{x}^{2}}-x-20 \right) \\
 & \Rightarrow y={{x}^{3}}-{{x}^{2}}-20x-3{{x}^{2}}+3x+60 \\
 & \therefore y={{x}^{3}}-4{{x}^{2}}-17x+60 \\
\end{align}$

Hence the cubic polynomial function in standard form with zeros 3,-4, and 5 is $y={{x}^{3}}-4{{x}^{2}}-17x+60$

Note: Cubic polynomial function is of type $f\left( x \right)=a{{x}^{3}}+b{{x}^{2}}+cx+d$ where $a\ne 0$. A cubic polynomial always has three zeros they can be all same or all different. The zeros can be real as well as complex depending on the function given. The zeros when substituted in the function given satisfied it and give the answer as zero. There is various methods to find the zeros of a cubic polynomial but the one commonly used have three step firstly we use hit and trial method and get one zero than using that zero factor form we divide our function and solve the quotient obtain using quadratic formula to get another two zeros.