Question

How do you find a polynomial function that has zeros $0,-2,-3$?

Answer 1

Solve for the required polynomial function:

As three zeros of a polynomial function are given, so the degree of the polynomial is $3$.

We know that a $3$ degree or cubic polynomial in terms of its factor is of the form $\mathrm{f}\left(\mathrm{x}\right)=k\left(\mathrm{x}-\mathrm{a}\right)\left(\mathrm{x}-\mathrm{b}\right)\left(\mathrm{x}-\mathrm{c}\right)$, where $\mathrm{a},\mathrm{b}$ and $\mathrm{c}$ are the zeros of the polynomial function.

So, a $3$ degree polynomial function with zeros $0,-2,-3$ can be obtained by substituting $\mathrm{a}=0,\mathrm{b}=-2$ and $\mathrm{c}=-3$ in the general form of cubic polynomial.

Thus, $\mathrm{f}\left(\mathrm{x}\right)=k\left(\mathrm{x}-0\right)\left(\mathrm{x}-\left(-2\right)\right)\left(\mathrm{x}-\left(-3\right)\right)$

$\Rightarrow$ $\mathrm{f}\left(\mathrm{x}\right)=k\mathrm{x}\left(\mathrm{x}+2\right)\left(\mathrm{x}+3\right)$

$\Rightarrow$ $\mathrm{f}\left(\mathrm{x}\right)=k\mathrm{x}\left({\mathrm{x}}^2+5x+6\right)$

$\Rightarrow$ $\mathrm{f}\left(\mathrm{x}\right)=k({\mathrm{x}}^3+5{\mathrm{x}}^2+6\mathrm{x})$

Hence, a polynomial function that has zeros $0,-2,-3$ is $\mathrm{k}({\mathrm{x}}^3+5{\mathrm{x}}^2+6\mathrm{x})$.