# How do you find all the zeros of #f(x) = x^3 - 4x^(2) + 14x - 20#?

##### 1 Answer

#### Explanation:

#f(x) = x^3-4x^2+14x-20#

By the rational root theorem, any rational zeros of

That means that the only possible rational zeros are:

#+-1# ,#+-2# ,#+-4# ,#+-5# ,#+-10# ,#+-20#

In addition note that if you invert the signs of the coefficients of the terms of odd degree, then the pattern of signs you get is:

#- - - -#

which has no changes. So there are no negative zeros.

So the only possible rational zeros are:

#1, 2, 4, 5, 10, 20#

We find:

#f(2) = 8-16+28-20 = 0#

So

#x^3-4x^2+14x-20 = (x-2)(x^2-2x+10)#

The remaining quadratic factor has negative discriminant, so no Real zeros, but we can still factor it by completing the square and using the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with

#x^2-2x+10#

#=x^2-2x+1+3^2#

#=(x-1)^2-(3i)^2#

#=((x-1)-3i)((x-1)+3i)#

#=(x-1-3i)(x-1+3i)#

Hence Complex zeros:

#x = 1+-3i#