Chapter 2.8, Problem 7E

Solution

To Solve:

The polynomial inequality $\left(x+1\right){\left(x-3\right)}^2>0$ , by completing the factoring if needed and then using a sign chart.

To Graph:

Support the solution graphically.

  $x\in \left(-1,3\right)\cup \left(3,\infty \right)$

Given:

The polynomial inequality $\left(x+1\right){\left(x-3\right)}^2>0$ .

Concepts Used:

The graph of $y=f\left(x\right)$ where $f\left(x\right)$ is a polynomial in one variable crosses the $x$ -axis at the zeroes of $f\left(x\right)$ . It changes sign at a particular zero if the multiplicity of that zero is odd, it undergoes no such sign change if the multiplicity of that zero is even.

If ${\left(ax+b\right)}^n$ is a factor of the polynomial $f\left(x\right)$ then $-\frac{b}{a}$ is a zero of $f\left(x\right)$ with multiplicity $n$ .

Polynomials are continuous everywhere in $ℝ$ . So they can change signs only at the points where they cross the $x$ -axis.

The solution to $f\left(x\right)>0$are the region in the graph of $y=f\left(x\right)$ where the curve lies above the $x$ -axis. This represents the region where $f\left(x\right)$ is positive.

Calculations:

Let $f\left(x\right)=\left(x+1\right){\left(x-3\right)}^2$ , the zeroes of $f\left(x\right)$ are $-3$ with an even multiplicity $2$ and $-1$ with an odd multiplicity $1$ . Thus, $x=-1$ will yield a sign change and $x=3$ will not.

Draw the sign chart based on this analysis:

  

  $\left(x+1\right){\left(x-3\right)}^2$ is positive when $x\in \left(-1,3\right)\cup \left(3,\infty \right)$ .

Graph:

Draw the graph of $y=f\left(x\right)$ and shade the region where it lies above the $x$ -axis.

  

The shaded area is the same as the analytically obtained solution.

Conclusion:

The solution of $\left(x+1\right){\left(x-3\right)}^2>0$ is $\left(3,\infty \right)$ .