Question 42199
To avoid any error in this type of problems follow the following procedure.
{{{P(x)= 16x^4-81}}}
={{{(4x^2)^2-9^2}}}
= {{{(4x^2+9)(4x^2-9)}}}
= {{{(4x^2+9)((2x)^2-3^2)}}}
= {{{(4x^2+9)(2x+3)(2x-3)}}}


This expression cannot be factorized as the formula is {{{a^2-b^2=(a+b)(a-b)}}} and not {{{a^2+b^2=(a+b)(a-b)}}}.
Here you have commited error.


However, {{{4x^2+9}}} can be factorized further using the concept of Complex Numbers.
There you will find {{{i=sqrt(-1)}}} i.e. 'i' is the imaginery square-root of {{{-1}}}.
So, {{{i^2=-1}}}.
Multiplying both side by {{{-9}}},
{{{-9i^2=9}}} i.e. {{{9=-9i^2}}}


Hence, {{{4x^2+9}}} 
= {{{4x^2-9i^2}}}
= {{{(2x)^2-(3i)^2}}}
= {{{(2x+3i)(2x-3i)}}}


Thus, {{{P(x) = (2x+3i)(2x-3i)(2x+3)(2x-3)}}}.
So the values of 'x' for which P(x) = 0 are: x = {{{-3i/2}}}, {{{3i/2}}}, {{{-3/2}}} and {{{3/2}}}