Method Three: Solve x2 + 2x - 8 ≤ 0 algebraically

x2 + 2x - 8 ≤ 0

( x + 4 )( x - 2 ) ≤ 0


Since ( x + 4 )( x - 2 ) ≤ 0, either     ( x + 4 ) ≥ 0 AND ( x - 2 ) ≤ 0     OR     ( x + 4 ) ≤ 0 AND ( x - 2 ) ≥ 0



( + )( - ) ≤ 0     OR     ( - )( + ) ≤ 0



[ x + 4 ≥ 0 AND x - 2 ≤ 0 ]     OR     [ x + 4 ≤ 0 AND x - 2 ≥ 0 ]        can be simplified to:


[ x ≥ -4 AND x ≤ 2 ]     OR     [ x ≤ -4 AND x ≥ 2 ]


[ x ≥ -4 AND x ≤ 2 ] can be re-written as [ -4 ≤ x AND x ≤ 2 ], then simplified to [ -4 ≤ x ≤ 2 ].

[ x ≤ -4 AND x ≥ 2 ] results in the empty set (Φ), since there are no numbers that are both less than or equal to -4 AND greater than or equal to 2


[ x ≥ -4 AND x ≤ 2 ]   OR   [ x ≤ -4 AND x ≥ 2 ]    =    [ -4 ≤ x ≤ 2 ]   OR   Φ


[ -4 ≤ x ≤ 2 ]   OR   Φ   =   -4 ≤ x ≤ 2


Solution Set
Solution set to the original inequality