How do you find the roots for #f(x) = 17x^15 + 41x^12 + 13x^3 - 10# using the fundamental theorem of algebra?

1 Answer
May 14, 2016

The FTOA does not help you find the zeros - it only tells you that this polynomial of degree #15# has exactly #15# Complex zeros counting multiplicity.

Explanation:

By roots, I will assume you mean zeros, i.e. values of #x# for which #f(x) = 0#.

The so called fundamental theorem of algebra (FTOA) is neither fundamental nor a theorem of algebra, but what it does tell you is that any non-zero polynomial in one variable with Complex coefficients has a zero in #CC#. That is, there is some Complex number #x_1# such that #f(x_1) = 0#.

A simple corollary of this - often stated as part of the FTOA - is that a polynomial in one variable of degree #n > 0# with Complex coefficients has #n# zeros counting multiplicity, all in #CC#.

In our example, #f(x)# is of degree #15#, so has #15# zeros counting multiplicity, all in #CC#.

The FTOA does not help you actually find the zeros.

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Bonus

What else can we find out about the zeros of this #f(x)#?

Note that the coefficients of #f(x)# have only one change of sign, so only one positive Real zero.

Note #f(0) = -10 < 0# and #f(1) = 17+41+13-10 = 61 > 0#

So the positive Real zero is in #(0, 1)#

#f(-x) = -17x^5+41x^12-13x^3-10#

has two changes of sign, so #f(x)# may have #0# or #2# negative Real zeros.

We find:

#f(-1) = -17+41-13-10 = 1 > 0#

So there is a Real zero in #(-1, 0)# and another Real zero in #(-oo, -1)#.

Any other zeros will occur in Complex conjugate pairs, since all of the coefficients of #f(x)# are Real.

Note that all of the degrees are multiples of #3# so the zeros are all cube roots of the zeros of:

#g(t) = 17t^5+41t^4+13t-10#