Problem 14 - College Algebra

Question

For the following exercises, solve the following polynomial equations by grouping and factoring. $$ 3 x^{3}-6 x^{2}-27 x+54=0 $$

Step-by-Step Solution

Answer

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

Step 1: Group the terms

The given polynomial equation is \(3x^3 - 6x^2 - 27x + 54 = 0\). To simplify, we'll group the terms in pairs: \((3x^3 - 6x^2)\) and \((-27x + 54)\).

Step 2: Factor out the common factors

Look at each group separately. For \(3x^3 - 6x^2\), factor out \(3x^2\), giving us \(3x^2(x - 2)\). For \(-27x + 54\), factor out \(-27\), giving us \(-27(x - 2)\).

Step 3: Identify the common binomial factor

Both factored expressions have a common binomial factor \((x - 2)\). Thus, the original equation becomes \(3x^2(x - 2) - 27(x - 2) = 0\).

Step 4: Factor by grouping

Since \((x - 2)\) is a common factor, it can be factored out of the equation: \((x - 2)(3x^2 - 27) = 0\).

Step 5: Solve each factor

Set each factor to zero: \(x - 2 = 0\) which gives \(x = 2\). Then solve \(3x^2 - 27 = 0\). Simplify to \(3x^2 = 27\) by adding 27 to both sides and dividing by 3, giving \(x^2 = 9\).

Step 6: Solve the quadratic equation

The equation \(x^2 = 9\) leads to two solutions: \(x = 3\) and \(x = -3\), since \(x = \pm\sqrt{9}\).

Step 7: List all solutions

The solutions to the polynomial equation \(3x^3 - 6x^2 - 27x + 54 = 0\) are \(x = 2\), \(x = 3\), and \(x = -3\).

Key Concepts

Factoring

Factoring is a crucial technique in solving polynomial equations. It involves breaking down an equation into simpler terms, called factors, that can be multiplied together to result in the original polynomial. Often, these factors are easier to manage and solve individually.

In our example, the polynomial equation is \(3x^3 - 6x^2 - 27x + 54 = 0\). The process begins with identifying parts of the equation that have common elements. For instance, if two terms both include \(x^2\), it might be possible to factor out \(x^2\).

To factor effectively, look for:

• Common numerical factors among coefficients.
• Common variable factors, like \(x^n\).
• Simple polynomial expressions that can be factored further, such as \(x^2 - 9\).

By factoring carefully, complex equations can become more approachable, allowing for easier calculation of possible solutions.

Grouping

Grouping is a strategy used to simplify polynomial equations. It involves rearranging and combining terms into groups with common factors. This can often make an otherwise complicated equation straightforward to solve.

In our example with the polynomial \(3x^3 - 6x^2 - 27x + 54 = 0\), grouping begins by pairing terms that can be factored together. Here, we separate them into \((3x^3 - 6x^2)\) and \((-27x + 54)\).

The key to successful grouping is:

• Identify distinct pairs or groups of terms.
• Ensure each group can be factored similarly.
• Find a common factor across the groups to enable further simplification.

Grouping is essentially a pre-factoring step that prepares the equation for easier solving. It can reveal hidden structures within the equation that simplify computations.

Quadratic Equation

A quadratic equation is a type of polynomial equation with a degree of two. It has the general form \(ax^2 + bx + c = 0\). Solving quadratic equations is an essential skill in algebra, often involving finding the roots of the equation.

In our example, part of the process involves solving the quadratic equation \(3x^2 - 27 = 0\) as part of the original polynomial factorization process. To find the roots, one approach is to isolate \(x^2\) by performing operations to simplify the equation:

1. Move the constant to the other side: \(3x^2 = 27\).
2. Divide by the leading coefficient (3): \(x^2 = 9\).
3. Take the square root of both sides, giving \(x = \pm3\).

By practicing these steps, students become adept at recognizing the structure of quadratics and applying appropriate techniques to find solutions efficiently.