Question

which graph shows the solution set of the inequality $\frac{x^{2}+10x + 16}{x - 3}>0$?

Explanation:

Step1: Factor the numerator

First, factor \(x^{2}+10x + 16=(x + 2)(x+8)\). So the inequality becomes \(\frac{(x + 2)(x + 8)}{x-3}>0\).

Step2: Find the critical - points

Set the numerator and denominator equal to zero. For the numerator \((x + 2)(x + 8)=0\), we get \(x=-2\) and \(x=-8\). For the denominator \(x - 3=0\), we get \(x = 3\). These are the critical - points: \(x=-8,x=-2,x = 3\).

Step3: Test the intervals

We have four intervals to test: \((-\infty,-8)\), \((-8,-2)\), \((-2,3)\) and \((3,\infty)\).

• For \(x<-8\), let \(x=-9\). Then \(\frac{(-9 + 2)(-9+8)}{-9-3}=\frac{(-7)(-1)}{-12}=-\frac{7}{12}<0\).
• For \(-8
• For \(-2
• For \(x>3\), let \(x=4\). Then \(\frac{(4 + 2)(4 + 8)}{4-3}=\frac{(6)(12)}{1}=72>0\).

The solution of the inequality \(\frac{x^{2}+10x + 16}{x-3}>0\) is the union of the intervals where the expression is positive, which is \((-8,-2)\cup(3,\infty)\). On a number - line, we have open circles at \(x=-8,x=-2,x = 3\) and the line is shaded for \(-83\).

Answer:

The graph that has open - circles at \(x=-8,x=-2,x = 3\) and is shaded for \(-83\) shows the solution set of the inequality. Without seeing the specific options, the solution set in interval notation is \((-8,-2)\cup(3,\infty)\).