Problem 38 - Precalculus : Building Concepts and Connections

Question

Find all real solutions of the polynomial equation. $$4 x^{3}-16 x^{2}+19 x=-6$$

Step-by-Step Solution

Answer

The real solutions of the polynomial equation are \(x = 0\) and \(x = 4\).

Step 1: Rearranging the Equation

Rearrange the given equation to the standard form. The standard form of a polynomial is ordered from highest to lowest degrees. Subtract -6 from both sides of the equation. The equation becomes: \(4x^3 - 16x^2 + 19x + 6 = 0\)

Step 2: Factoring by Grouping

This is a cubic equation so we can try to factor it by grouping. Break the polynomial into two sections: \((4x^3 - 16x^2) + (19x + 6)\). Now factor out a \(4x^2\) from the first group and a 1 from the second group: \(4x^2(x - 4) + 1(19x + 6)\).

Step 3: Solving for \(x\)

Set each factor equal to zero and solve for \(x\). This gives two possible solutions: \(4x^2 = 0\) gives \(x = 0\) and \(x - 4 = 0\) gives \(x = 4\). Note that the \(19x+6=0\) does not yield a real solution.

Key Concepts

Factoring by Grouping

Factoring by grouping is a powerful technique to simplify complex polynomial expressions. It's especially helpful when dealing with polynomial equations that don't lend themselves easily to simpler factoring methods. Here's how it works:

• First, split the polynomial into two groups. Each group should appear to have a common factor that you can factor out.
• In our example, the polynomial is \(4x^3 - 16x^2 + 19x + 6\). We can group this as \((4x^3 - 16x^2) + (19x + 6)\).
• Once you have your groups, factor out the greatest common factor in each group. For \(4x^3 - 16x^2\), we can factor out \(4x^2\), which gives us \(4x^2(x - 4)\). In \(19x + 6\), there isn't a common factor other than 1, so we write it as \(1(19x + 6)\).

After factoring out the common factors in each group, attempt to rearrange or further simplify to see if a solution can be identified by setting each factor to zero and solving for \(x\). This strategy often reveals potential solutions or roots of the equation.

Cubic Equations

Cubic equations are polynomial equations in which the highest power of the variable is three. These equations can often be more challenging to solve because they involve more complex algebra compared to quadratic equations. Thes are great strategies for tackling cubic equations, like:

• Rearranging the equation to standard polynomial form, where terms are ordered by the highest degree to lowest.
• Using methods like factoring by grouping, synthetic division, or sometimes even trial and error to factor the polynomial.

In our problem, the cubic equation in question is \(4x^3 - 16x^2 + 19x + 6 = 0\).
We approached it by factoring by grouping effectively to break it down into factors \(4x^2(x - 4) + 1(19x + 6)\). The goal is to set these to zero `\(x - 4 = 0\)`, `\(4x^2 = 0\)` to find solutions. Always check for further simplifications or factorizations in cubic equations to make solving easier.

Real Solutions

Real solutions are the root values of an equation that are real numbers. In polynomial equations, particularly higher-degree ones like cubic equations, identifying real solutions involves setting each factor obtained from factorization equal to zero and solving for \(x\).
In the exercise given, the process identifies potential real solutions by checking each factor:

• From the factor \(4x^2 = 0\), solving gives \(x = 0\).
• From the factor \(x - 4 = 0\), solving gives \(x = 4\).
• The term \(19x + 6\) doesn’t yield a real solution as it doesn’t simplify to a zero. It implies no intersection with the x-axis for real numbers.

Real solutions are where the function equals zero on a graph, meaning they are also x-intercepts of the equation's graph. It's important to confirm that factors are indeed zero when evaluating and disregard any which do not yield valid real numbers. This keeps your answer valid and precise.