How do you find the number of roots for #f(x) = 2x^3 – 12x^2 + 11x + 2# using the fundamental theorem of algebra?

1 Answer
Nov 25, 2016

The fundamental theorem of algebra tells us that this cubic has #3# zeros counting multiplicity.

We can further discover that all #3# are Real with #2# distinct positive zeros and one negative.

Explanation:

#f(x) = 2x^3-12x^2+11x+2#

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

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

This follows since given a polynomial of degree #n > 0# with a zero #x=a#, we can divide the polynomial by #(x-a)# to give a polynomial of degree #n-1#. If #n - 1 > 0# then it has a possibly distinct zero #b#, etc.

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

We can find out more by examining the discriminant #Delta#, which for a cubic of the form #ax^3+bx^2+cx+d# is given by the formula:

#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#

In our example, #a=2#, #b=-12#, #c=11# and #d=2#, so we find:

#Delta = 17424-10648+13824-432-9504 = 10664#

Since #Delta > 0#, all three zeros of #f(x)# are Real and distinct.

Using Descartes's Rule of Signs we can also tell that #2# of these zeros are positive and one negative.