Answer
$(2p^{2}+3)(3p^{2}-1)$
Work Step by Step
Substitute $u=p^{2}$
$6p^{4}+7p^{2}-3=6u^{2}+7u-3=...$
When factoring $ax^{2}+bx+c$, we search for factors of $ac$ whose sum is $b,$
and, if we find them, we rewrite $bx$ and proceed to factor in groups.
Here, factors of $(6)\times(-3)=-18$ that add to $+7$ are ....$+9$ and $-2$
$6u^{2}+7u-3=6u^{2}+9u-2u-3=$
$=3u(2u+3)-(2u+3)$
$=(2u+3)(3u-1)$
... bring back p
$=(2p^{2}+3)(3p^{2}-1)$
