How do you solve #x^3-9x^2+14x=48# using the fundamental theorem of algebra?

1 Answer
Jul 14, 2016

The FTOA only tells you that it has #3# roots, but we can find by other means that they are:

#8# and #1/2+-sqrt(23)/2i#

Explanation:

We want to find the zeros of:

#f(x) = x^3-9x^2+14x-48#

The fundamental theorem of algebra only tells us that since #f(x)# is of degree #3# then it has exactly #3# Complex (possibly Real) zeros counting multiplicity. It does not tell us how to find them.

We can use the rational root theorem, which tells us that any rational zeros of #f(x)# can be expressed in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-48# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1, +-2, +-3, +-4, +-6, +-8, +-12, +-16, +-24, +-48#

Trying each in turn, we eventually find:

#f(8) = 512-576+112-48 = 0#

So #x=8# is a zero and #(x-8)# a factor:

#x^3-9x^2+14x-48 = (x-8)(x^2-x+6)#

The remaining quadratic has negative discriminant, so only non-Real Complex zeros, but we can still use the quadratic formula to find them:

#x = (1+-sqrt((-1)^2-4(1)(6)))/(2*1) = 1/2+-sqrt(23)/2i#