How do you find the number of roots for #0=x^3-4x-48# using the fundamental theorem of algebra?

1 Answer
Dec 9, 2015

The degree of this polynomial is #3# so by the fundamental theorem of algebra it has #3# roots (counting multiplicity).

Explanation:

A non-zero polynomial equation in one variable always has as many roots as its degree. They may be repeated and/or Complex, but it will have that many roots.

The highest degree term in #0 = x^3-4x-48# is the #x^3# one, so the polynomial is of degree #3#.

Let #f(x) = x^3-4x-48#.

Notice that #f(4) = 64-16-48 = 0#, so #x=4# is one of the roots and #(x-4)# is a factor of #f(x)#.

#x^3-4x-48 = (x-4)(x^2+4x+12)#

The remaining quadratic factor #(x^2+4x+12)# is of the form #ax^2+bx+c#, with #a=1#, #b=4# and #c=12#. It has zeros given by the quadratic formula:

#x=(-b+-sqrt(b^2-4ac))/(2a) = (-4+-sqrt((-4)^2-(4xx1xx12)))/(2xx1)#

#=(-4+-sqrt(-32))/2 = -2+-2sqrt(2) i#

..that is a conjugate pair of Complex roots.