# How do you find an example of a fourth-degree polynomial equation that has no real zeros?

Jun 6, 2016

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#### Explanation:

Note that if a polynomial has Real coefficients, then any non-Real Complex zeros occur in Complex conjugate pairs.

So to construct a quartic with no Real zeros, start with two pairs of Complex conjugate numbers.

For example, we could pick $1 \pm i$ and $2 \pm i$,

Then multiply out:

$f \left(x\right) = \left(x - \left(1 + i\right)\right) \left(x - \left(1 - i\right)\right) \left(x - \left(2 + i\right)\right) \left(x - \left(2 - i\right)\right)$

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

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

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

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

$= {x}^{4} - 6 {x}^{3} + 15 {x}^{2} - 18 x + 10$

Or you could simply start with any quartic polynomial with positive leading coefficient, then increase the constant term until it no longer intersects the $x$ axis.