# How do you write a polynomial function of least degree and leading coefficient 1 when the zeros are -2, -3, i, -i?

Aug 13, 2017

$P \left(x\right) = \left(x + 2\right) \left(x + 3\right) \left({x}^{2} + 1\right)$
$\text{ } = {x}^{4} + 5 {x}^{3} + 7 {x}^{2} + 5 x + 6$

#### Explanation:

Suppose the polynomial is $P \left(x\right)$

By the factor theorem, if $x = a$ is root of $P \left(x\right) = 0$, then $x = a$ is a factor of $P \left(x\right)$

We have the following roots of $P \left(x\right) = 0$

$x = - 2 , - 3 , i , - i$

Hence, the following are factors of $P \left(x\right)$

$\left(x + 2\right) , \left(x + 3\right) , \left(x - i\right) , \left(x + i\right)$

Hencde, we can write the polynoimal of least degree as the product of these factors (any higher degree polynomial would have additional roots)

$P \left(x\right) = A \left(x + 2\right) \left(x + 3\right) \left(x - i\right) \left(x + i\right)$

We want our polynomial to have leading coefficient $1 \implies A = 1$, and also to have real coefficients, so let us multiply out the complex factors (as they are conjugates we will get a real product)

$\left(x - i\right) \left(x + i\right) = {x}^{2} + i x - i x - {i}^{2} = {x}^{2} + 1$

Thus we have:

$P \left(x\right) = \left(x + 2\right) \left(x + 3\right) \left({x}^{2} + 1\right)$
$\text{ } = {x}^{4} + 5 {x}^{3} + 7 {x}^{2} + 5 x + 6$