Answer
$(A\cup B)=\{x:-4\le x\le 2\}$
$(A\cap B)=\{x:-3\le x\le 2\}$
Work Step by Step
Let A be the set of x’s for which ${{x}^{2}}+2x\le 8$; let B be the set for which ${{x}^{2}}+x\le 6$
Since the first equation factors into $\left( x+4 \right)\left( x-2 \right)\le 0$, its solution set is \[A=\left\{ x:-4\le x\le 2 \right\}\] Similarly, the second equation can be written $\left( x+3 \right)\left( x-2 \right)\le 0$, making \[B=\left\{ x:-3\le x\le 2 \right\}\].
Therefore,
$(A\cap B)=\{x:-3\le x\le 2\}$
and
$(A\cup B)=\{x:-4\le x\le 2\}$
