#### Answer

$\frac{(h-1)(h+4)}{3}$

#### Work Step by Step

$\frac{h-1}{6h+3}\times(2h^2+9h+4)$
$=\frac{h-1}{3(2h+1)}\times(2h^2+8h+h+4)$
$=\frac{h-1}{3(2h+1)}\times[2h(h+4)+(h+4)]$
$=\frac{h-1}{3(2h+1)}\times(2h+1)(h+4)$
$=\frac{(h-1)(2h+1)(h+4)}{3(2h+1)}$
$=\frac{(h-1)(h+4)}{3}$