# How do you write a polynomial function of least degree with integral coefficients that has the given zeros 3, 2, -1?

$y = \left(x - 3\right) \left(x - 2\right) \left(x + 1\right)$
Also
$y = {x}^{3} - 4 {x}^{2} + x + 6$

#### Explanation:

From the given zeros 3, 2, -1

We set up equations $x = 3$ and $x = 2$ and $x = - 1$. Use all these as factors equal to the variable y.

Let the factors be $x - 3 = 0$ and $x - 2 = 0$ and $x + 1 = 0$

$y = \left(x - 3\right) \left(x - 2\right) \left(x + 1\right)$

Expanding

$y = \left({x}^{2} - 5 x + 6\right) \left(x + 1\right)$
$y = \left({x}^{3} - 5 {x}^{2} + 6 x + {x}^{2} - 5 x + 6\right)$
$y = {x}^{3} - 4 {x}^{2} + x + 6$

Kindly see the graph of $y = {x}^{3} - 4 {x}^{2} + x + 6$ with zeros at $x = 3$ and $x = 2$ and $x = - 1$

God bless....I hope the explanation is useful.