How do you write a polynomial function of minimum degree with real coefficients whose zeros include those listed: 5i and √5?

Nov 8, 2015

$f \left(x\right) = \left({x}^{2} + 25\right) \left(x - \sqrt{5}\right) = {x}^{3} - \sqrt{5} {x}^{2} + 25 - 25 \sqrt{5}$

or if you want rational coefficients:

$g \left(x\right) = \left({x}^{2} + 25\right) \left({x}^{2} - 5\right) = {x}^{4} + 20 {x}^{2} - 125$

Explanation:

If a polynomial has Real coefficients, then any non-Real Complex roots will occur in Complex conjugate pairs. So the roots of our polynomial must include $5 i$, $- 5 i$ and $\sqrt{5}$.

If we allow irrational coefficients then the monic polynomial of lowest degree with these roots is:

$f \left(x\right) = \left(x - 5 i\right) \left(x + 5 i\right) \left(x - \sqrt{5}\right) = \left({x}^{2} + 25\right) \left(x - \sqrt{5}\right)$

$= {x}^{3} - \sqrt{5} {x}^{2} + 25 - 25 \sqrt{5}$

To have rational coefficient then we also need the irrational conjugate $- \sqrt{5}$ of $\sqrt{5}$ resulting in:

$g \left(x\right) = \left(x - 5 i\right) \left(x + 5 i\right) \left(x - \sqrt{5}\right) \left(x + \sqrt{5}\right) = \left({x}^{2} + 25\right) \left({x}^{2} - 5\right)$

$= {x}^{4} + 20 {x}^{2} - 125$

Any polynomial with these roots will be a multiple (scalar or polynomial) of $f \left(x\right)$ or $g \left(x\right)$