According to the fundamental theorem of algebra, how many roots does the polynomial #f(x)=x^4+3x^2+7# have over the complex numbers, and counting roots with multiplicity greater than one as distinct? (i.e #f(x)=x^2# has two roots, both are zero).

1 Answer
George C.
Dec 14, 2017

#4#

Explanation:

The fundamental theorem of algebra (FTOA) tells us that a polynomial of degree #n>0# with complex (possibly real) coefficients has a zero in the complex numbers.

A straightforward corollary of the FTOA, often stated as part of it, is that a polynomial of degree #n>0# with complex (possibly real) coefficients has exactly #n# complex (possibly real) zeros counting multiplicity.

In our example we are given:

#f(x) = x^4+3x^2+7#

which is of degree #4#.

Hence by the FTOA (and corollary) it has exactly #4# zeros counting multiplicity.

Bonus: Find the zeros

Note that:

#t^2+3t+7#

has discriminant:

#Delta = color(blue)(3)^2-4(color(blue)(1))(color(blue)(7)) = 9-28=-19#

Since #Delta < 0# this quadratic has a complex conjugate pair of non-real zeros.

So attempting to solve #x^4+3x^2+7=0# as a quadratic in #x^2# then taking the square roots would give us square roots of complex numbers to simplify.

Let's use another approach:

Since #f(x)# has no terms of odd degree, it will factor in the form:

#x^4+3x^2+7 = (x^2-kx+sqrt(7))(x^2+kx+sqrt(7))#

#color(white)(x^4+3x^2+7) = x^4+(2sqrt(7)-k^2)x^2+7#

Putting:

#2sqrt(7)-k^2 = 3#

we find:

#k^2 = 2sqrt(7)-3#

So:

#k = +-sqrt(2sqrt(7)-3)#

So:

#x^4+3x^2+7 = (x^2-sqrt(2sqrt(7)-3)x+sqrt(7))(x^2+sqrt(2sqrt(7)-3)x+sqrt(7))#

Hence zeros given by the quadratic formula:

#x = 1/2(+-sqrt(2sqrt(7)-3)+-sqrt(2sqrt(7)+3)i)#

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