Answer
$-(3x+1)(3x-1)(x^2+1)$
Work Step by Step
Let $x^2=a$.
$1-8x^2-9x^4=1-8a-9a^2$
Rewrite $-8a$ as $-9a+1a$
$=1-9a+1a-9a^2$
Group the first two terms together and group the last two terms together.
$=(1-9a)+(1a-9a^2)$
Factor out the GCF in each group.
$=1(1-9a)+a(1-9a)$
Factor out $(1-9a)$.
$=(1-9a)(1+a)$
Back substitute $a=x^2$.
$=(1-9x^2)(1+x^2)$
$=[1^2-(3x)^2](1+x^2)$
Use special formula $a^2-b^2=(a+b)(a-b)$where $a=1$ and $b=3x$.
$=(1+3x)(1-3x)(1+x^2)$
$=(1+3x)[-(-1+3x)](1+x^2)
$=(3x+1)[-(3x-1)](x^2+1)
$=-(3x+1)(3x-1)(x^2+1)$
Hence, the completely factored form of the given expression is $-(3x+1)(3x-1)(x^2+1)$.
