How do you find the roots for #f(x) = 15x^12 + 41x^9 + 13x^2 –10# using the fundamental theorem of algebra?

1 Answer
Jun 5, 2016

The FTOA tells us it has #12# zeros but does not help us find them...

Explanation:

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

A corollary of this, often stated as part of the FTOA is that a polynomial of degree #n > 0# in one variable will have exactly #n# Complex (possibly Real) zeros counting multiplicity.

To see why this follows, note that if we have one zero #x_1#, then #f(x)# will be divisible by #(x-x_1)# resulting in a polynomial of degree #n - 1#. If #n - 1 > 0# then this will have a zero #x_2#, etc.

In our example:

#f(x) = 15x^12+41x^9+13x^2-10#

is of degree #12#, so has exactly #12# zeros counting multiplicity.

That's all the FTOA tells you. It does not help you find the zeros.

#color(white)()#
What else can we find out about the zeros of this polynomial?

Note that the signs of the coefficients follow the pattern:

#+ + + -#

With one change of sign, that means that there will be exactly one positive Real zero.

Reversing the sign on the term of odd degree we get the pattern:

#+ - + -#

With three changes of sign, that means that there will be either #1# or #3# negative Real zeros.

We can try to find rational zeros using the rational root theorem:

Any zero of #f(x)# must be expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-10# and #q# a divisor of the coefficient #15# of the leading term.

That means that the only possible rational zeros are:

#+-1/15#, #+-2/15#, #+-1/5#, #+-1/3#, #+-2/5#, #+-2/3# #+-1#, #+-5/3#, #+-2#, #+-5#, #+-10#

None of these works, so #f(x)# has no rational zeros.

Ultimately, we are stuck with using numerical methods (e.g. Durand-Kerner) to find approximations for the zeros:

#x~~-1.38923#

#x~~0.735661#

#x~~-0.775742+-0.203247i#

#x~~-0.487897+-0.736584i#

#x~~0.14955+-0.928279i#

#x~~0.714459+-1.21625i#

#x~~0.726412+-0.459408i#