# How do you write a polynomial with zeros 3-i, 5i and leading coefficient 1?

Oct 23, 2016

${x}^{4} - 6 {x}^{3} + 35 {x}^{2} - 150 x + 250 = 0$

#### Explanation:

For polynomial equations with real coefficients, complex roots occur

in conjugate pairs. so, the least-degree polynomial equation having

roots $3 \pm i \mathmr{and} \pm 5 i$ is

$\left(x - 3 - i\right) \left(x - 3 + i\right) \left(x - 5 i\right) \left(x + 5 i\right) = 0$ that simplifies to

$\left({\left(x - 3\right)}^{2} + 1\right) \left({x}^{2} + 25\right) = 0$ and this expands to

${x}^{4} - 6 {x}^{3} + 35 {x}^{2} - 150 x + 250 = 0$

.