How do you find all the zeros of #y=x^2+8x+15# with its multiplicities?

1 Answer
Oct 30, 2016

Answer:

The zeros are located at #x=-5# and #x=-3#. They both have a multiplicity of #1#.

Explanation:

Find the zeros and multiplicities of the zeros of
#y=x^2+8x+15#

Factor

#y=(x+5)(x+3)#

Set each factor to zero and solve.

#x+5=0# and #x+3=0#

#x=-5# and #x=-3#

The zeros are located at #x=-5# and #x=-3#.

They each have a multiplicity of #color(red)1# because the exponent of each factor is #color(red)1# as seen below.

#y=(x+5)^color(red)1(x+3)^color(red)1#