Answer
$(3t^{2}+5)(4t^{2}-7)$
Work Step by Step
Substitute $u=t^{2}.$
$12t^{4}-t^{2}-35=12u^{2}-u-35=...$
When factoring $ax^{2}+bx+c$,
we search for factors of $ac=-12\cdot 35=-420$ whose sum is $b=-1,$
Here, these factors are $-21$ and $+20$
$...=12u^{2}+20u-21u-35=$
$=4u(3u+5)-7(3u+5)$
$=(3u+5)(4u-7)$
... bring back $t$
$=(3t^{2}+5)(4t^{2}-7)$
