How do you find the zeroes for #y=(x-3)^3(x+1)#?

1 Answer
George C.
Jun 7, 2015

#y = (x-3)^3(x+1) = (x-3)(x-3)(x-3)(x+1)#

This will be zero when any of the linear factors are zero, that is when #x=-1# or #x=3#.

If all of the linear factors are non-zero, then #y# will not be zero either.

This is a quartic - a polynomial of order #4# - in #x#. All quartic equations in one variable have a total of #4# roots, but some may be complex numbers or repeated. In your case, there is one root #x=-1# of multiplicity #1# and one repeated root #x=3# of multiplicity #3#.

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