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

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Answer 1 · 2016-10-30T03:07:11

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

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#