The product ( x − 2 ) ( x + 3 ) is positive if both factors are negative or if both factors are positive. Therefore, we can solve ( x − 2 ) ( x + 3 ) as follows, ( x − 2 < and x + 3 < 0 ) ( x < 2 and x < − 3 ) x < − 3 o r o r o r ( x − 2 > 0 and x + 3 > 0 ) ( x > 2 and x > − 3 ) x < 2 The solution set is ( − ∞ , − 3 ) ∪ ( 2 , ∞ ) . Use this type of analysis to solve each of the following. (a) ( x − 2 ) ( x + 7 ) > 0 (b) ( x − 3 ) ( x + 9 ) ≥ 0 (c) ( x + 1 ) ( x − 6 ) ≤ 0 (d) ( x + 4 ) ( x − 8 ) < 0 (e) x + 4 x − 7 > 0 (f) x + 4 x − 7 > 0

The product ( x − 2 ) ( x + 3 ) is positive if both factors are negative or if both factors are positive. Therefore, we can solve ( x − 2 ) ( x + 3 ) as follows,

( x − 2 <   and   x + 3 < 0 ) ( x < 2   and   x < − 3 ) x < − 3 o r o r o r ( x − 2 > 0   and   x + 3 > 0 ) ( x > 2   and   x > − 3 ) x < 2

The solution set is ( − ∞ , − 3 ) ∪ ( 2 , ∞ ) . Use this type of analysis to solve each of the following.

(a) ( x − 2 ) ( x + 7 ) > 0	(b) ( x − 3 ) ( x + 9 ) ≥ 0
(c) ( x + 1 ) ( x − 6 ) ≤ 0	(d) ( x + 4 ) ( x − 8 ) < 0
(e) x + 4 x − 7 > 0	(f) x + 4 x − 7 > 0