## Precalculus (6th Edition)

$(2m^{2}+9)(5m^{2}-1)$
Substitute $u=m^{2}.$ $10m^{4}+43m^{2}-9=10u^{2}+43u-9=...$ When factoring $ax^{2}+bx+c$, we search for factors of $ac=-90$ whose sum is $b=43,$ Here, these factors are $-2$ and $+45$ $...=10u^{2}+45u-2u-9$ $=5u(2u+9)-(2u+9)$ $=(2u+9)(5u-1)$ ... bring back $m$ $=(2m^{2}+9)(5m^{2}-1)$