What is all of the real and imaginary zeros of #y=(x^2-9)(x^2+9)(x+3)^2#?

Question

What is each zeroes' multiplicity?

438 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-12T11:38:59

We have #4# zeros, #3# with multiplicity #3# and #-3,3i# and #-3i# with multiplicity of #1#.

#y=(x^2-9)(x^2+9)(x-3)^2#

= #(x^2-3^2)(x^2-(3i)^2)(x-3)^2#

= #(x+3)(x-3)(x-3i)(x+3i)(x-3)^2#

= #(x-(-3))(x-3)(x-3i)(x-(-3i))(x-3)^2#

= #(x-(-3))(x-3i)(x-(-3i))(x-3)^3#

Hence, we have #4# zeros, #3# with multiplicity #3# and #-3,3i# and #-3i# with multiplicity of #1#.

Answer 2 · 2017-12-12T11:53:24

Real zeros are #x=-3# with multiplicity #3# and #x=3# ,Imaginary zeros are #x=-3i# and #x=3i#

#y=(x^2-9)(x^2+9)(x+3)^2#

#(x^2-9)=0 or (x+3)(x-3)=0:. x =-3 , x=3 #

#(x^2+9)=0 or (x^2-(9i^2)=0 [i^2=-1]# or

#x^2-(3i)^2=0 or (x+3i)(x-3i)=0# or

#:. x =-3i , x=3i #

#(x+3)^2=0 or (x+3)(x+3)=0 :. x=-3 , x=-3#

Zeros are # x= (-3,3,-3i,3i,-3 ,-3)#

Real zeros are #x=-3# with multiplicity #3# and #x=3#

Imaginary zeros are #x=-3i# with and #x=3i# [Ans]