# How do you write the polynomial function of least degree with integral coefficients that has the give zeros for 2, 4i?

Nov 25, 2016

$f \left(x\right) = {x}^{3} - 2 {x}^{2} + 16 x - 32$

#### Explanation:

If the function has a zero at $4 i$, it also has one at $- 4 i$.

If a function has a zero at $a$, it has a factor of $x - a$.

So, this function has factors of $\left(x - 2\right)$, $\left(x - 4 i\right)$, and $\left(x + 4 i\right)$.

The function can be written as

$f \left(x\right) = \left(x - 2\right) \left(x - 4 i\right) \left(x + 4 i\right)$

Mutliplying $\left(x - 4 i\right) \left(x + 4 i\right)$ gives

$f \left(x\right) = \left(x - 2\right) \left({x}^{2} + 4 i - 4 i - 16 {i}^{2}\right)$

Recall that ${i}^{2} = - 1$

$f \left(x\right) = \left(x - 2\right) \left({x}^{2} + 16\right)$

Multiply each term in the first binomial by each term in the second binomial.

$f \left(x\right) = {x}^{3} + 16 x - 2 {x}^{2} - 32$

Rearranging....

$f \left(x\right) = {x}^{3} - 2 {x}^{2} + 16 x - 32$