Answer
$\displaystyle \frac{2(100x^{2}+95x+11)}{(4x+3)^{2}}$
Work Step by Step
$\left[ (4x+3)^{-1} =(4x+3)^{-2+1}=(4x-1)(4x-1)^{-2} \right]$
$(4x+3)^{-1}-8(4x+3)^{-2}+10(5x+1)(4x+3)^{-1}=$
$= (4x+3)^{-1}-8(4x+3)^{-2}+10(5x+1)(4x-1)(4x-1)^{-2}$
... factor out $2(4x+3)^{-2}$
$=2(4x+3)^{-2}[-4+5(5x+1)(4x+3)]$
$=2(4x+3)^{-2}[-4+5(20x^{2}+19x+3)]$
$=2(4x+3)^{-2}[-4+100x^{2}+95x+15)] \qquad $... apply $a^{-n}=\displaystyle \frac{1}{a^{n}}$
$=\displaystyle \frac{2(100x^{2}+95x+11)}{(4x+3)^{2}}$
