Question

Evaluate :
$39y^3({50y}^2-98)÷16y^2(5y+7)$

Answer 1

$39y^3(50y^2-98)\times$ reciprocal of $16y^2(5y+7)$

$=\frac{39y^3(50y^2-98)}{16y^2(5y+7)}$

$=\frac{39y^3\times 2(25y^2-49)}{16y^2(5y+7)}$

$=\frac{78y(5y-7)(5y+7)}{16(5y+7)}$

$=\frac{39y(5y-7)}{8}$