Jan 17, 2018

$- 5 \cdot {x}^{3} + 12$ is one example.

#### Explanation:

Since the degree must be 3 we know there will be an $a \cdot {x}^{3}$ term.

Since there must be a negative leading coefficient, we can make $a$ any negative number, so let's say $- 5 \cdot {x}^{3}$.

We need a binomial, which means there will be two terms, and the constant term must greater than, so we'll add a constant to what we currently have:

$- 5 \cdot {x}^{3} + 12$, for example.

That 12 could actually be any positive number, so $- 5 \cdot {x}^{3} + 85$ would also work.