Solve the nonlinear inequality: (
x 1)(x+3)(x 7)2 0 Solve by marking zeros and undefined points on a number line and checking signs in the intervals created as in the examples in the narrative. Express your answer in interval notation.

Solve the nonlinear inequality: (
x 1)(x+3)(x 7)2 0 Solve by marking zeros and undefined points on a number line and checking signs in the intervals created as in the examples in the narrative. Express your answer in interval notation.

A product is greater zero if the number of negative signs is even.

Since $(x-7)^2 > 0~\implies~x \in \mathbb{R} \setminus \{7\}$ you only have to examine the remaining 2 factors:

$\underbrace{x-1\geq 0\wedge x+3 \geq 0}_{\text{both factors positive}}~\vee~ \underbrace{x-1< 0\wedge x+3 < 0}_{\text{both factors negative}}$

After some simplification you'll get:

$x \in (-\infty,-3] \cup [1,\infty)$

3. ## Thank you - can you help with this???

Solve the inequality:
x
(x + 2)(x 5)(x + 7) 0 Solve by marking zeros and undefined points
on a number line and checking signs in the intervals created as in the examples in the narrative. Express your answer in interval notation.

Solve the inequality: . $x(x+ 2)(x - 5) (x + 7)\:\leq\: 0$

Solve by marking zeros and undened points on a number line
and checking signs in the intervals created.
The zeros of the inequality are: . $-7,\:-2,\:0,\:5$

These four values divide the number line into five intervals.

. . $\begin{array}{cccccccccc}
\underbrace{\;-----} & o & \underbrace{----- }& o & \underbrace{-----} & o & \underbrace{-----} & o & \underbrace{-----\;} \\
& \text{-}7 && \text{-}2 && 0 && 5 \end{array}$

Now test a value from each interval:

On $(\text{-}\infty, \text{-}7)$, try $x = \text{-}8:\;\;(\text{-}8)(\text{-}8+2)(\text{-}8-5)(\text{-}8+7) \:=\:(\text{-}8)(\text{-}6)(\text{-}13)(\text{-}1) \:=\:+182$ . . . no

On $(\text{-}7,\text{-}2)$, try $x = \text{-}3\!:;\;(\text{-}3)(\text{-}3+2)(\text{-}3-5)(\text{-}3+7) \:=\:(\text{-}3)(\text{-}1)(\text{-}8)(4) \:=\:-96$ . . . yes

On $(\text{-}2,0)$, try $x = \text{-}1\!:\;\;(\text{-}1)(\text{-}1+2)(\text{-}1-5)(\text{-}1+7) \:=\:(\text{-}1)(1)(\text{-}6)(6) \:=\:+36$ . . . no

On $(0,5)$, try $x = 1\!:\;\;(1)(1+7)(1-5)(1+7) \:=\:(1)(8)(\text{-}6)(6) \:=\:-288$ . . . yes

On $(5,\infty)$, try $x = 6\!:\;\;(6)(6+2)(6-5)(6+7) \:=\:(6)(8)(1)(13) \:=\:+624$ . . . no

The inequality is satisfied on: . $(\text{-}7,\text{-}2) \,\cup\, (0,5)$