Problem 42 - College Algebra with Modeling and Visualization

Question

Solve the polynomial inequality (a) symbolically and (b) graphically. $$ 8 x^{3}<27 $$

Step-by-Step Solution

Answer

The solution to the inequality \( 8x^3 < 27 \) is \( x < \frac{3}{2} \).

Step 1: Rewrite the Inequality in Standard Form

Start by rewriting the inequality in the form: \( 8x^3 - 27 < 0 \). This helps us to identify key components and factors.

Step 2: Factor the Polynomial

Recognize that \( 8x^3 - 27 \) is a difference of cubes. It can be factored using the identity \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \). Let \( a = 2x \) and \( b = 3 \), so the inequality becomes \( (2x - 3)((2x)^2 + (2x)(3) + 3^2) < 0 \). Simplifying, we get \( (2x - 3)(4x^2 + 6x + 9) < 0 \).

Step 3: Find the Critical Points

Set each factor equal to zero to find the critical points: \( 2x - 3 = 0 \) gives \( x = \frac{3}{2} \). The second factor, \( 4x^2 + 6x + 9 \), does not factor further, and its discriminant \( b^2 - 4ac = 36 - 144 = -108 \) is negative, so it has no real roots.

Step 4: Analyze and Determine Intervals

Since only \( x = \frac{3}{2} \) is a real critical point, check intervals determined by this point. Test intervals: \((-\infty, \frac{3}{2})\) and \((\frac{3}{2}, \infty)\). Choose test points from each interval, e.g., \( x = 0 \) for \((-\infty, \frac{3}{2})\) and \( x = 2 \) for \((\frac{3}{2}, \infty)\). Evaluate \( 8x^3 - 27 \) at these points to check sign.

Step 5: Evaluate Test Points

Evaluate at \( x = 0: \, 8(0)^3 - 27 = -27 \), which is less than 0. Evaluate at \( x = 2: \, 8(2)^3 - 27 = 64 - 27 = 37 \), which is greater than 0.

Step 6: Solve Graphically (Optional)

Graph the function \( f(x) = 8x^3 - 27 \) using a graphing calculator or tool. Identify where the graph is below the x-axis, confirming solutions are for \( x < \frac{3}{2} \).

Step 7: Determine the Solution

Combining findings from symbolic and graphical solutions, the solution is \( x < \frac{3}{2} \), as this interval keeps \( 8x^3 - 27 \) negative.

Key Concepts

Polynomial Factoring

Factoring plays a crucial role when dealing with polynomial inequalities. Let's take a closer look at the inequality \( 8x^3 - 27 < 0 \). We need to rewrite it in a form that makes it easier to analyze.

When you see something like \( 8x^3 - 27 \), think about special factoring formulas. This is a special kind called a difference of cubes. The formula for factoring a difference of cubes is \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \).

Here it goes like this:

• Let \( a = 2x \) and \( b = 3 \).
• Using the formula, \( 8x^3 - 27 \) turns into \( (2x - 3)(4x^2 + 6x + 9) \).

Now, the polynomial is separated into factors we can work with to solve the inequality. This step is crucial because it simplifies the expression into parts we can easily manage.

Critical Points

Critical points in a polynomial inequality are what help us determine where and how the inequality changes. After factoring the polynomial \( 8x^3 - 27 \) into \( (2x - 3)(4x^2 + 6x + 9) < 0 \), we need to find where each part equals zero.

Critical points are found by setting each factor equal to zero:

• For \( 2x - 3 = 0 \), solving gives us \( x = \frac{3}{2} \).
• The other factor \( 4x^2 + 6x + 9 \) has no real roots because its discriminant is negative.

A discriminant tells us about the roots of a quadratic. If it's negative, the quadratic doesn't cross the x-axis, meaning no real roots. Therefore, the only critical point to consider here is \( x = \frac{3}{2} \). It segments the x-axis into different intervals for further analysis.

Graphical Solution

Using a graphical solution gives a visual representation of the inequality \( 8x^3 - 27 < 0 \). By graphing the function \( f(x) = 8x^3 - 27 \), we can visually see where the function is below the x-axis. This is where the inequality holds true.

You can use a graphing calculator or software to plot the function. Look closely at the curve:

• The section of the graph that lies below the x-axis represents where \( f(x) \) is less than zero.
• Notice where the curve crosses the x-axis.

In this problem, you will notice that \( x = \frac{3}{2} \) is the point of interest.

The portion of the graph left of this point is below the x-axis. This confirms that the solution to the inequality is for \( x < \frac{3}{2} \). Graphical solutions are handy to verify the correct signs found in symbolic solutions.

Interval Analysis

Interval analysis involves examining different sections of the number line to understand how the factored polynomial behaves in each section. With \( 8x^3 - 27 < 0 \) factored to \( (2x - 3)(4x^2 + 6x + 9) < 0 \), and \( x = \frac{3}{2} \) determined as the critical point, we can split the number line into intervals.

The main intervals to consider are:\((-\infty, \frac{3}{2})\) and \((\frac{3}{2}, \infty)\).

From here, use a test point from each interval to see if the inequality holds:

• Choose \( x = 0 \) for \((-\infty, \frac{3}{2})\). Plugging in, \( 8(0)^3 - 27 = -27 \), which is less than 0.
• Choose \( x = 2 \) for \((\frac{3}{2}, \infty)\). Here, \( 8(2)^3 - 27 = 64 - 27 = 37 \), which is greater than 0.

This test indicates that the solution \( x < \frac{3}{2} \) is valid because the polynomial is negative in that interval. Interval analysis pairs nicely with other methods, confirming solutions by checking different regions on the number line.