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

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Answer 1 · 2016-08-11T23:03:16

#f(x)# has zeros #1#, #3# and #4# all with multiplicity #1#.

#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 #f(1) = 0#, #x=1# is a zero and #(x-1)# a factor:

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

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

To factor the remaining quadratic, note that #3+4=7# and #3xx4=12#.

So we find:

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

with associated zeros #x=3# and #x=4#.