How do you find all zeros with multiplicities of #f(x)=x^3-2x^2-5x+6#?

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Answer 1 · 2017-08-22T19:21:27

The zeros of #f(x)# are #1, 3, -2#, each with multiplicity #1#.

Given:

#f(x) = x^3-2x^2-5x+6#

Note that the sum of the coefficients is #0#. That is:

#1-2-5+6 = 0#

Hence we find #f(1) = 0# and #(x-1)# is a factor:

#x^3-2x^2-5x+6 = (x-1)(x^2-x-6)#

To factor the remaining quadratic find a pair of factors of #6# which differ by #1#. The pair #3, 2# works, so we find:

#x^2-x-6 = (x-3)(x+2)#

So the zeros of #f(x)# are #1, 3, -2#, each with multiplicity #1#.