How do you find the number of roots for #f(x) = 3x^4 + x + 2# using the fundamental theorem of algebra?

1 Answer
George C.
Dec 7, 2016

By the FTOA, #f(x)# has exactly #4# Complex, possibly Real zeros counting multiplicity.

Further we find that none are Real.

Explanation:

The fundamental theorem of algebra (FTOA) tells us that any polynomial in one variable of degree #n > 0# has a Complex (possibly Real) zero.

A straightforward corollary of this, often stated as part of the FTOA is that a polynomial of degree #n > 0# has exactly #n# Complex, possibly Real, zeros counting multiplicity.

In our example:

#f(x) = 3x^4+x+2#

is of degree #4# and therefore has #4# Complex, possibly Real, zeros counting multiplicity.

What else can we find out about these zeros?

The pattern of signs of the coefficients of #f(x)# is #+ + +#. With no changes of sign, we can deduce that #f(x)# has no positive Real zeros by Descartes' Rule of Signs.

The pattern of signs of #f(-x)# is #+ - +#. With #2# changes of sign, it means that #f(x)# has #0# or #2# negative Real zeros.

Note that when #abs(x) <= 1#, we have #3x^4 >= 0# and since #abs(x) < 2#, we find:

#f(x) > 0#

When #abs(x) > 1#, then #abs(3x^4) = abs(3x^3 * x) > 3 abs(x)#
Hence:

#f(x) > 0#

So #f(x)# has no Real zeros. All of its #4# zeros must be non-Real Complex.

Related questions
See all questions in Fundamental Theorem of Algebra
Impact of this question
3100 views around the world
You can reuse this answer
Creative Commons License