Answer
$\frac{y-3}{y+6}$.
Work Step by Step
The given expression is
$\frac{y^2+4y-21}{y^2+3y-28}\div \frac{y^2+14y+48}{y^2+4y-32}$
$\frac{y^2+4y-21}{y^2+3y-28}\cdot \frac{y^2+4y-32}{y^2+14y+48}$
Factor each denominator and numerator.
Steps to factor
1. Rewrite the middle term.
2.Group terms.
3. Factor from each group.
4. Factor out common terms.
The factors are:-
$=y^2+4y-21$
$=y^2+7y-3y-21$
$=(y^2+7y)+(-3y-21)$
$=y(y+7)-3(y+7)$
$=(y+7)(y-3)$
$=y^2+3y-28$
$=y^2+7y-4y-28$
$=(y^2+7y)+(-4y-28)$
$=y(y+7)-4(y+7)$
$=(y+7)(y-4)$
$=y^2+4y-32$
$=y^2+8y-4y-32$
$=(y^2+8y)+(-4y-32)$
$=y(y+8)-4(y+8)$
$=(y+8)(y-4)$
$=y^2+14y+48$
$=y^2+8y+6y+48$
$=(y^2+8y)+(6y+48)$
$=y(y+8)+6(y+8)$
$=(y+8)(y+6)$
Substitute all factors into the given expression.
$=\frac{(y+7)(y-3)}{(y+7)(y-4)}\cdot \frac{(y+8)(y-4)}{(y+8)(y+6)}$
Cancel common factors.
The remaining fraction is
$=\frac{y-3}{y+6}$.
