How do you find a polynomial function that has the given zeros. 0, –3?

Nov 5, 2015

${x}^{2} + 3 x$

Explanation:

We know that if a product $a \cdot b = 0$, then either $a = 0$ or $b = 0$ (or both). So, if we want a polynomial to have certain zeroes, in this case $0$ and $- 3$, we can multiply to polynomials that have those zeroes.

It's clear that the choice is not unique, but we usually choose the easiest ones: if we want ${x}_{0}$ to be a zero of the polynomial, $\left(x - {x}_{0}\right)$ is surely a good option, since the result in ${x}_{0}$ gives ${x}_{0} - {x}_{0} = 0$.

So, if we want $0$ to be a solution, our polynomial would be $\left(x - 0\right) = x$. As for $- 3$, the same steps lead us to $\left(x - \left(x - 3\right)\right) = \left(x + 3\right)$

Now we have a polynomial with a root in zero (namely $x$), and a polynomial with a root in $- 3$ (namely $x + 3$). If we multiply them, we have the desired polynomial:

$x \left(x + 3\right) = 0$ if $x = 0$ or $x + 3 = 0$, i.e. $x = - 3$.

To write it as an explicit polynomial, simply expand the expression:

$x \left(x + 3\right) = {x}^{2} + 3 x$