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

Algebra ->  Algebra  -> Polynomials-and-rational-expressions -> 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      Log On

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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(489) About Me  (Show Source):
You can put this solution on YOUR website!
To avoid any error in this type of problems follow the following procedure.
P%28x%29=+16x%5E4-81
=%284x%5E2%29%5E2-9%5E2
= %284x%5E2%2B9%29%284x%5E2-9%29
= %284x%5E2%2B9%29%28%282x%29%5E2-3%5E2%29
= %284x%5E2%2B9%29%282x%2B3%29%282x-3%29

This expression cannot be factorized as the formula is a%5E2-b%5E2=%28a%2Bb%29%28a-b%29 and not a%5E2%2Bb%5E2=%28a%2Bb%29%28a-b%29.
Here you have commited error.

However, 4x%5E2%2B9 can be factorized further using the concept of Complex Numbers.
There you will find i=sqrt%28-1%29 i.e. 'i' is the imaginery square-root of -1.
So, i%5E2=-1.
Multiplying both side by -9,
-9i%5E2=9 i.e. 9=-9i%5E2

Hence, 4x%5E2%2B9
= 4x%5E2-9i%5E2
= %282x%29%5E2-%283i%29%5E2
= %282x%2B3i%29%282x-3i%29

Thus, P%28x%29+=+%282x%2B3i%29%282x-3i%29%282x%2B3%29%282x-3%29.
So the values of 'x' for which P(x) = 0 are: x = -3i%2F2, 3i%2F2, -3%2F2 and 3%2F2