# 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#