# How do you find all the zeros of #f(x)= x^3 - 8x^2 +19x - 12# with its multiplicities?

##### 1 Answer

Aug 11, 2016

#### Explanation:

#f(x) = x^3-8x^2+19x-12#

First note that the sum of the coefficients is zero. That is:

#1-8+19-12 = 0#

Hence

#x^3-8x^2+19x-12#

#=(x-1)(x^2-7x+12)#

To factor the remaining quadratic, note that

So we find:

#x^2-7x+12 = (x-3)(x-4)#

with associated zeros