# How do you write a polynomial function of least degree and leading coefficient 1 when the zeros are -6, 3, 5?

Jul 24, 2016

$f \left(x\right) = \left(x + 6\right) \left(x - 3\right) \left(x - 5\right) = {x}^{3} - 2 {x}^{2} - 33 x + 90$

#### Explanation:

For each zero $a$, the polynomial must have a corresponding linear factor $\left(x - a\right)$.

So the polynomial of least degree with all three zeros is:

$f \left(x\right) = \left(x + 6\right) \left(x - 3\right) \left(x - 5\right) = {x}^{3} - 2 {x}^{2} - 33 x + 90$

Any polynomial in $x$ with these zeros is a multiple (scalar or polynomial) of this $f \left(x\right)$.