How do you find all zeros with multiplicities of #f(x)=6x^4-5x^3-9x^2#?

1 Answer
Alvin L.
Dec 20, 2017

#x=0# (multiplicity #2#) and #x=(5+-sqrt241)/12# (multiplicity #1# for both)

Explanation:

First we can factor out an #x^2#:
#6x^4-5x^3-9x^2=x^2(6x^2-5x-9)#

From this, it's quite clear that #x=0# is a solution, and it will have multiplicity #2# because the factor is #x^2#.

To find the remaining solutions, we can use the quadratic formula to solve:
#6x^2-5x-9=0#

#x=(5+-sqrt241)/12#

These solutions only occur once, so they both have a multiplicity of #1#.

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