How do you find all the zeros of #P(x) = (x-5)^2(x+2)^3(x+4)#?

1 Answer
May 28, 2016

Answer:

Note that #x# is a zero if and only it causes at least one of the factors of #P(x)# to be zero. Hence zeros:

#5# (multiplicity #2#)

#-2# (multiplicity #3#)

#-4#

Explanation:

#P(x) = (x-5)^2(x+2)^3(x+4)#

#=(x-5)(x-5)(x+2)(x+2)(x+2)(x+4)#

Note that #P(x) = 0# if and only if at least one of its factors is zero.

So the only zeros are:

  • #x=5# with multiplicity #2#
  • #x=-2# with multiplicity #3#
  • #x=-4# with multiplicity #1#

Essentially, finding the zeros of a polynomial (with their multiplicities) is the same as finding the linear factors of that polynomial. If #(x-a)# is a factor then #x=a# is a zero and vice versa.