How do you find the number of roots for #F(x)=x^3-10x^2+27x-12# using the fundamental theorem of algebra?

1 Answer
Feb 28, 2016

The FTOA allows us to infer that #F(x) = 0# has #3# roots.

Further analysis allows us to find the #3# Real roots that it has.

Explanation:

The fundamental theorem of algebra tells you that any non-constant polynomial with Complex (possibly Real) coefficients has a zero in the Complex numbers.

A simple corollary of this is that a polynomial of degree #n > 0# will have #n# zeros in the Complex numbers, counting multiplicity.

To show the corollary, notice that if a polynomial #f(z)# of degree #n > 0# has a zero #r_1 in CC# then the polynomial is divisible by #(z-r_1)# to give a polynomial of degree #n-1#. If #n-1 > 0# then this simpler polynomial will have a zero #r_2 in CC# (which may be equal to #r_1#). We can then divide the simpler polynomial by #(z-r_2)#, etc. Repeat this process until you find:

#f(z) = a(z-r_1)(z-r_2)...(z-r_n)#

where #r_1, r_2,..., r_n in CC#

In our example #F(x)# is of degree #3#, so has exactly #3# Complex (possibly Real) zeros counting multiplicity.

What else can we find out about these roots of #F(x) = 0# ?

Notice that the coefficients of #F(x)# have #3# changes of sign. So #F(x)# may have up to #3# positive Real zeros.

On the other hand, #F(-x)# has no changes of sign, so #F(x)# has no negative Real zeros.

Note that since #F(x)# has Real coefficients, any non-Real Complex roots must occur in Complex conjugate pairs.

Hence we find that #F(x) = 0# either has #3# positive Real roots or #1# positive Real roots and a pair of non-Real Complex conjugate roots.

By the rational root theorem, any rational roots must be expressible in the form #p/q# for integers #p#, #q# with #p# a divisor of the constant term #-12# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational roots are:

#1, 2, 3, 4, 6, 12#

(positive since we know there are no negative zeros).

Trying each of these in turn we find #F(4) = 0#. So #x=4# is a root and #(x-4)# a factor:

#x^3-10x^2+27x-12 = (x-4)(x^2-6x+3)#

We can solve the remaining quadratic factor using the quadratic formula to find roots:

#x = (6+-sqrt(24))/2 = 3+-sqrt(6)#

So there are #3# positive Real roots.