How do you solve #x^3-9x^2+14x=48# using the fundamental theorem of algebra?
1 Answer
The FTOA only tells you that it has
#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
We can use the rational root theorem, which tells us that any rational zeros of
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^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#