Solve: $\frac { x ^ { 2 } - 2 x - 24 } { x + 7 } 0$
A) $( - 7 , - 4 ) \cup ( 6 , \infty )$
B) $( - \infty , - 7 ) \cup ( - 4,6 )$
C) $( - \infty , \infty )$
D) $( - 2 , - 12 )$

Question

Solve: $\frac { x ^ { 2 } - 2 x - 24 } { x + 7 } > 0$
A) $( - 7 , - 4 ) \cup ( 6 , \infty )$
B) $( - \infty , - 7 ) \cup ( - 4,6 )$
C) $( - \infty , \infty )$
D) $( - 2 , - 12 )$

Quizplus · Accepted Answer

First, factor the numerator to get $(x-6)(x+4)$. The inequality becomes $\frac{(x-6)(x+4)}{x+7} > 0$. The critical points are $x = -7, -4, 6$. Test values in each interval to determine where the expression is positive. The solution is $x \in (-7, -4) \cup (6, \infty)$.

Solve x2−2x−24x+7>0\frac { x ^ { 2 } - 2 x - 24 } { x + 7 } > 0x+7x2−2x−24​>0

Solve $\frac { x ^ { 2 } - 2 x - 24 } { x + 7 } > 0$