SOLUTION: Factor the polynomial P(x)= 16x^4-81 completely and find all its zeros. I factored to (4x^2+9)(4x^2-9) then further to ((2x+3)(2x-3))^2 to get 3/2 and -3/2 both with multiplic

Question 42199This question is from textbook
: Factor the polynomial P(x)= 16x^4-81 completely and find all its zeros.
I factored to (4x^2+9)(4x^2-9) then further to ((2x+3)(2x-3))^2 to get 3/2 and -3/2 both with multiplicity of 2 to equal four zeros. But the solutions given also contain 3i/2 and -3i/2. Also, I'm unsure that the second factoring step is appropriate since by distribution (2x+3)^2 is not equal to 4x^2+9 etc. I'm stumped! Thanks for any help! This question is from textbook

Answer by psbhowmick(878) (Show Source): You can put this solution on YOUR website!
To avoid any error in this type of problems follow the following procedure.

=
=
=
=

This expression cannot be factorized as the formula is and not .
Here you have commited error.

However, can be factorized further using the concept of Complex Numbers.
There you will find i.e. 'i' is the imaginery square-root of .
So, .
Multiplying both side by ,
i.e.

Hence,
=
=
=

Thus, .
So the values of 'x' for which P(x) = 0 are: x = , , and
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