Problem 29 - Precalculus: Functions and Graphs

Question

Solve the equation. \(9 x^{3}-18 x^{2}-4 x+8=0\)

Step-by-Step Solution

Answer

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

Step 1: Factor by grouping

First, we need to try factoring the polynomial by grouping. Split the equation into two groups: \[ (9x^3 - 18x^2) + (-4x + 8) = 0 \]Factor out the greatest common factor (GCF) from each group. From the first group, factor out \(9x^2\), and from the second group, factor out \(-4\): \[ 9x^2(x - 2) - 4(x - 2) = 0 \]

Step 2: Factor out the common factor

Notice that both terms in the equation have a common factor \((x - 2)\). Factor \((x - 2)\) out:\[ (x - 2)(9x^2 - 4) = 0 \]

Step 3: Solve for the roots

Use the zero-product property, which states if \(ab = 0\), then \(a = 0\) or \(b = 0\). So we have two equations to solve: \[ x - 2 = 0 \] \[ 9x^2 - 4 = 0 \]First, solve the linear equation:\[ x - 2 = 0 \rightarrow x = 2 \]

Step 4: Solve the quadratic equation

Next, solve the quadratic equation for \(x\):\[ 9x^2 - 4 = 0 \]Add 4 to both sides:\[ 9x^2 = 4 \]Divide by 9:\[ x^2 = \frac{4}{9} \]Take the square root of both sides:\[ x = \pm \frac{2}{3} \]

Step 5: List the solutions

The solutions to the equation \(9x^3 - 18x^2 - 4x + 8 = 0\) are: \[ x = 2, \quad x = \frac{2}{3}, \quad x = -\frac{2}{3} \]

Key Concepts

Factoring by Grouping

Factoring by grouping is a useful method when you need to simplify polynomial equations, especially those with four terms. It allows you to break down more complex polynomials into manageable pieces. You start by dividing the polynomial into two separate groups. For example, consider the polynomial \(9x^3 - 18x^2 - 4x + 8\). Here, you group it as \((9x^3 - 18x^2)\) and \((-4x + 8)\). Then, factor out the greatest common factor (GCF) in each group:

• For \(9x^3 - 18x^2\), the GCF is \(9x^2\), so you get \(9x^2(x - 2)\).
• For \(-4x + 8\), the GCF is \(-4\), resulting in \(-4(x - 2)\).

This step will lead you to find a common factor, making further factoring and simplification much easier. The whole expression simplifies to \((9x^2 - 4)(x - 2)\). This approach makes it easier to find solutions to the equation.

Zero-Product Property

The zero-product property is an essential principle in algebra used to find the solutions to equations. It states that if a product of two factors is zero, then at least one of the factors must be zero. This concept applies beautifully once a polynomial equation is factored into simpler binomials or trinomials. Let's look at our example:

• Starting from the factorized form \((9x^2 - 4)(x - 2) = 0\), apply the zero-product property.
• Set each factor equal to zero: \(9x^2 - 4 = 0\) and \(x - 2 = 0\).

Solving these equations individually gives the roots of the original polynomial. For the linear equation \(x - 2 = 0\), you immediately find \(x = 2\). Solving the quadratic \(9x^2 - 4 = 0\) requires further steps, but the zero-product property has reduced a complex problem to simpler sub-problems.

Quadratic Equations

Quadratic equations are polynomials of degree two. Solving them is a fundamental skill in algebra. They typically appear in the form \(ax^2 + bx + c = 0\). Here in our exercise, when factoring by grouping, one of the results was the quadratic \(9x^2 - 4 = 0\). You can solve this using several methods, but in this case, let's isolate \(x\) by direct manipulation:

• Add \(4\) to both sides of the equation, yielding \(9x^2 = 4\).
• Divide by \(9\) to get \(x^2 = \frac{4}{9}\).
• Take the square root of both sides, remembering the \(\pm\) sign for squaring, leading to \(x = \pm \frac{2}{3}\).

This mathematical approach reveals two solutions for \(x\), \(x = \frac{2}{3}\) and \(x = -\frac{2}{3}\). Knowing how to manipulate and solve these equations is vital for tackling more complex models in mathematics.