How do you find the number of roots for #0=x^3+450x^2-1350# using the fundamental theorem of algebra?

1 Answer
Dec 28, 2015

This polynomial equation is of degree #3#, so has exactly #3# Complex roots counting multiplicity, some of which may be Real.

Actually all three roots are Real and distinct.

Explanation:

The fundamental theorem of algebra tells you that a polynomial equation in one variable of degree #n > 0# has at least one Complex root.

A simple corollary of this is that a polynomial equation in one variable of degree #n > 0# has #n# possibly repeated Complex roots. This is sometimes included as part of the statement of the fundamental theorem of algebra.

Here's a sketch of the proof of the corollary:

If a polynomial #f(x)# has one Complex zero #x_1#, then #(x-x_1)# is a factor of #f(x)# and you can write #f(x) = (x-x_1)g(x)# where #g(x)# is of degree #n-1#. Then if #n-1 > 0#, #g(x) = 0# has at least one Complex root #x_2#, which might be the same as #x_1#, etc. So counting multiplicity, a polynomial of degree #n# has exactly #n# Complex roots.

Note that the Complex roots referred to in all of this may be Real or non-Real.

The equation #0 = x^3+450x^2-1350# is of degree #3#, hence has #3# Complex roots counting multiplicity. That's all the fundamental theorem of algebra will tell you.

Beyond this, since this polynomial has Real coefficients, any non-Real Complex roots will occur in conjugate pairs. So we can deduce that our polynomial will have at least one Real root.

It so happens that all of these #3# roots are Real in this example, being approximately #-450#, #-1.74# and #1.73#

To show that all of the roots are Real, let #f(x) = x^3+450x^2-1350# and evaluate #f(x)# for a few values:

#f(-450) = -450^3+450^3-1350 = -1350 < 0#

#f(-2) = -8+450*4-1350 = -8+1800-1350 = 442 > 0#

#f(0) = 0 + 0 - 1350 = -1350 < 0#

#f(2) = 8+450*4-1350 = 8+1800-1350 = 458 > 0#

So #f(x)# crosses the #x# axis #3# times and #f(x) = 0# has #3# distinct Real roots.