Question

Solve your answers using inequalities using a number line strategy or a factor table strategy. Express your answers using set notation. a) (x+3)^2 (x+1)(x-1) > 0b) (x+1)(x+3) ≥ 0

Answer 1

$S=\lbrace x\parallel x1\rbrace$

Explanation

$(x+3)^2(x+1)(x-1)>0$

Step 1

solve for each factor:

set each factor equal to zero

$\begin{gathered} (x+3)^2=0 \\ x=-3 \end{gathered}$$\begin{gathered} x+1=0 \\ x=-1 \end{gathered}$$\begin{gathered} x-1=0 \\ x=1 \end{gathered}$

so, we have 3 critical numbers

-3,-1 and 1

Step 2

now, let's make a table to evaluate each interval , (select a number from the interval)

$\begin{gathered} a)(x+3)^2 \\ \text{for x=-5} \\ (x+3)^2=(-5+3)^2=4 \\ so\text{ }\Rightarrow positive \\ \text{for x= -2} \\ (x+3)^2=5^2=25\Rightarrow positive \\ \text{for x= 0} \\ (0+3)^2=9\Rightarrow positive \\ for\text{ x= 2} \\ (2+3)^2=(2+3)^2=25\Rightarrow positive \end{gathered}$

b) (x+1)

$\begin{gathered} x+1 \\ \text{for x=-4} \\ -4+1=-3\Rightarrow negative \\ \text{for x=-2} \\ -2+1=-1\Rightarrow negative \\ \text{for x=0} \\ 0+1=1\Rightarrow positive \\ \text{for x= 2} \\ 2+1=3\Rightarrow positive \end{gathered}$

c)(x-1)

$\begin{gathered} x-1 \\ \text{for x=-4} \\ -4-1=-\Rightarrow negative \\ \text{for x=-2} \\ -2-1=-3\Rightarrow negative \\ \text{for x=0} \\ 0-1=-1\Rightarrow\text{negative} \\ \text{for x= 2} \\ 2-1=1\Rightarrow positive \end{gathered}$

Step 3

now, make th multiplication of the signs ( check the table in step 2)

so, we got

(-infi, -1) U (1, inf)

as we need that the product of the factors is greater than zero ( positive) , the solution is( set notation)

$S=\lbrace x\parallel x1\rbrace$

I hope this helps you