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

1 Answer
George C.
Jun 6, 2016

See explanation...

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+-i# and #2+-i#,

Then multiply out:

#f(x) = (x-(1+i))(x-(1-i))(x-(2+i))(x-(2-i))#

#=((x-1)-i)((x-1)+i)((x-2)-i)((x-2)+i)#

#=((x-1)^2-i^2)((x-2)^2-i^2)#

#=((x-1)^2+1)((x-2)^2+1)#

#=(x^2-2x+2)(x^2-4x+5)#

#=x^4-6x^3+15x^2-18x+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.

Related questions
See all questions in Zeros
Impact of this question
18591 views around the world
You can reuse this answer
Creative Commons License