How do you find all zeros with multiplicities of #f(x)=x^3+4x^2-11x+6#?

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Answer 1 · 2017-03-04T01:14:44

The zeros of #f(x)# are #1# with multiplicity #2# and #-6# with multiplicity #1#.

Given:

#f(x) = x^3+4x^2-11x+6#

First note that the sum of the coefficients is #0#. That is:

#1+4-11+6 = 0#

Hence #x=1# is a zero and #(x-1)# a factor:

#x^3+4x^2-11x+6 = (x-1)(x^2+5x-6)#

The sum of the coefficients of the remaining quadratic is also #0#, so #x=1# is a zero again and #(x-1)# a factor:

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

So the remaining zero is #x=-6#.

So the zeros of #f(x)# are #1# with multiplicity #2# and #-6# with multiplicity #1#.