Problem 16 - Precalculus: Functions and Graphs

Question

Find all solutions of the equation. $$x^{3}+x^{2}-14 x-24=0$$

Step-by-Step Solution

Answer

The solutions are \(x = -2, 4, -3\).

Step 1: Identify Possible Rational Roots

Using the Rational Root Theorem, potential rational roots are the factors of the constant term (-24) divided by the leading coefficient (1). Thus, potential roots are: ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24.

Step 2: Test Rational Roots Using Synthetic Division

Perform synthetic division on the polynomial using each potential root to find one that results in a remainder of 0.Trying -2, we perform synthetic division on the polynomial:\[\begin{array}{c|cccc}-2 & 1 & 1 & -14 & -24 \ & & -2 & 2 & 24 \\hline & 1 & -1 & -12 & 0 \\end{array}\]The remainder is 0, indicating that \(x = -2\) is a root.

Step 3: Factor the Polynomial

Since \(x = -2\) is a root, the polynomial \(x^3 + x^2 - 14x - 24\) factors into \((x + 2)(x^2 - x - 12)\).

Step 4: Factor the Quadratic

The quadratic \(x^2 - x - 12\) can be factored further. We look for two numbers whose product is -12 and sum is -1. These numbers are -4 and 3.Thus, \(x^2 - x - 12 = (x - 4)(x + 3)\).

Step 5: Write the Complete Factored Form

The complete factorization of the polynomial is \((x + 2)(x - 4)(x + 3) = 0\).

Step 6: Solve for the Roots

Set each factor equal to zero and solve for \(x\):1. \(x + 2 = 0\) gives \(x = -2\).2. \(x - 4 = 0\) gives \(x = 4\).3. \(x + 3 = 0\) gives \(x = -3\).

Key Concepts

Rational Root Theorem

In solving polynomial equations, the Rational Root Theorem is a valuable tool. It helps us identify possible rational roots of a polynomial equation. If you have a polynomial equation with integer coefficients, the potential rational roots can be found by taking the factors of the constant term and dividing them by the factors of the leading coefficient. For our equation, \(x^3 + x^2 - 14x - 24 = 0\), the constant term is \(-24\) and the leading coefficient is \(1\). Therefore, the potential rational roots could be \(\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24\).

By testing these potential roots, we can determine which, if any, are actual roots of the polynomial. This reduces the number of solutions to check manually, speeding up the solving process. It's important to note that not all potential roots listed by the theorem are guaranteed to be actual roots.

Synthetic Division

After listing the potential rational roots, the next step is to test these values using synthetic division. This method is a simplified form of polynomial long division. It allows us to evaluate whether a potential root is an actual root.

To perform synthetic division, we align the coefficients of the polynomial and the potential root. Then, we carry out a series of multiplications and additions. If the remainder at the end of the division is zero, the candidate is indeed a root. In our example, when \(-2\) was tested, the synthetic division output a remainder of zero, affirming that \(x = -2\) is a root.

Using synthetic division reduces errors and saves time compared to longer traditional division methods.

Factoring Polynomials

Once you have found a root using synthetic division, you can factor the polynomial further. Factoring involves rewriting the polynomial as a product of its factors. Since we determined that \(x = -2\) is a root, we can express the polynomial \(x^3 + x^2 - 14x - 24\) as \((x + 2)(x^2 - x - 12)\).

The factoring process simplifies the remaining polynomial, making it easier to solve. This further break down can eventually lead us to find all roots of the polynomial equation. Remember, factoring polynomials completely is crucial to understanding the solutions to the equation.

Quadratic Equations

After factoring the polynomial and isolating the quadratic part, the next step is to solve the resulting quadratic equation. A quadratic equation is generally in the form \(ax^2 + bx + c = 0\). For our factor \(x^2 - x - 12\), we need to find numbers that multiply to get \(-12\) and add up to get \(-1\).

The numbers \(-4\) and \(3\) satisfy these conditions, leading us to factor the quadratic as \((x - 4)(x + 3)\). Solving each factor by setting it to zero gives the solutions \(x = 4\) and \(x = -3\).

Tackling quadratics after initial factoring simplifies the solving process of polynomial equations. Always double-check your factors and solutions to ensure they fit the original equation.