Problem 12 - A Graphical Approach to Precalculus with Limits

Question

Find all real solutions. Do not use a calculator. $$3 x^{3}+18=x(2 x+27)$$

Step-by-Step Solution

Answer

The real solutions are \(x = 3, x = -3,\) and \(x = \frac{2}{3}\).

Step 1: Expand Both Sides

Start by expanding the right-hand side of the given equation. We are starting with the equation:\[3x^3 + 18 = x(2x + 27)\]Expanding the right-hand side gives:\[2x^2 + 27x\]So, the equation becomes:\[3x^3 + 18 = 2x^2 + 27x\]

Step 2: Combine Like Terms

Next, we need to rearrange terms so that one side of the equation is zero. Move all terms to the left side:\[3x^3 + 18 - 2x^2 - 27x = 0\]This simplifies to:\[3x^3 - 2x^2 - 27x + 18 = 0\]

Step 3: Factor by Grouping

To factor this cubic polynomial, we can first consider grouping terms:\[3x^3 - 2x^2 - 27x + 18\]Group the terms as follows:\[(3x^3 - 2x^2) - (27x - 18)\]Factor out common factors from each group:\[x^2(3x - 2) - 9(3x - 2)\]

Step 4: Factor Completely

Since both groups contain the common factor \((3x - 2)\), we can factor this out:\[(x^2 - 9)(3x - 2) = 0\]Notice that \(x^2 - 9\) is a difference of squares, which can be factored further:\[(x - 3)(x + 3)(3x - 2) = 0\]

Step 5: Solve Each Factor

Set each factor equal to zero and solve for \(x\):\[x - 3 = 0 \rightarrow x = 3\]\[x + 3 = 0 \rightarrow x = -3\]\[3x - 2 = 0 \rightarrow 3x = 2 \rightarrow x = \frac{2}{3}\]

Step 6: Conclusion

The solutions found are the values of \(x\) that satisfy the original equation. Therefore, the real solutions are:\[x = 3, x = -3, \text{ and } x = \frac{2}{3}\]

Key Concepts

Factoring

Factoring is a hugely beneficial tool in algebra, especially when dealing with polynomial equations. It involves expressing a polynomial as a product of simpler polynomials, which can make solving equations far easier.
One common approach is to use **factoring by grouping**, which splits the polynomial into groups that can be factored separately. For instance, in the polynomial equation \(3x^3 - 2x^2 - 27x + 18 = 0\), first, it's divided into two groups:

• \(3x^3 - 2x^2\)
• \(-27x + 18\)

From the first group, you can factor out \(x^2\), and from the second group, \(-9\). This results in:

• \(x^2(3x - 2) - 9(3x - 2)\)

Notice the common factor \((3x - 2)\), which allows further factoring:

• \((x^2 - 9)(3x - 2) = 0\)

Remember, recognizing patterns like the difference of squares, which in this case \(x^2 - 9\) is equal to \((x - 3)(x + 3)\), simplifies factoring further.

Cubic Polynomials

Cubic polynomials can appear complex at first but understanding their components makes them manageable. A cubic polynomial is one where the highest exponent of the variable is three, such as in \(3x^3 - 2x^2 - 27x + 18\). They can be solved by factoring, as shown in this exercise.
At the core, solving cubic polynomials often involves finding solutions to equations like \(ax^3 + bx^2 + cx + d = 0\). Methods such as "factoring by grouping" are immensely useful because they break complex expressions into manageable parts.
Once factors are determined, like in our example \((x^2 - 9)(3x - 2)\), you transform the cubic equation into simpler quadratic or linear forms. Thus, cubic polynomials often resolve into a set of simpler equations, ready to be tackled through factoring or using other algebraic techniques.

Real Solutions

In the context of polynomial equations, real solutions are the values of \(x\) that satisfy the equation and lie on the real number line. For the polynomial \((x - 3)(x + 3)(3x - 2) = 0\), finding real solutions involves setting each factor to zero and solving for \(x\).
This exercise illustrates it well:

• \(x - 3 = 0\) leads to \(x = 3\)
• \(x + 3 = 0\) gives \(x = -3\)
• \(3x - 2 = 0\) solves to \(x = \frac{2}{3}\)

These values, \(x = 3, x = -3, \text{ and } x = \frac{2}{3}\), are the real solutions because they fulfill the original cubic polynomial equation without requiring imaginary or complex numbers.
Discovering real solutions is critical in algebra as it’s often the end goal of solving polynomial equations. It confirms that each equation is balanced by the values found. Some polynomials might have no real solutions, but in this example, all solutions are real.