Answer
$(x^{2}+5)(7x^{2}-1)$
Work Step by Step
$7x^{4}+34x^{2}-5$
Temporarily substitute $t=x^{2}$.
The expression is now
$7t^{2}+34t-5=$
For the trinomial $ax^{2}+bx+c$, we search for factors of $ac$ whose sum is $b$. We find $+35(-1)=7\cdot(-5),\quad 35-1=+34$
Rewrite bx and factor in pairs
$=7t^{2}+35t-t-5$
$=7t(t+5)-(t+5)$
$=(t+5)(7t-1)$
bring back x
$=(x^{2}+5)(7x^{2}-1)$
