Problem 15 - Precalculus: Functions and Graphs

Question

Find all solutions of the equation. $$x^{3}-x^{2}-10 x-8=0$$

Step-by-Step Solution

Answer

The solutions are \(x = -1, \: x = 4, \: x = -2.\)

Step 1: Identify Possible Rational Roots

To find the rational roots, we use the Rational Root Theorem, which suggests that any rational solution is of the form \(\frac{p}{q}\), where \(p\) is a factor of the constant term (-8) and \(q\) is a factor of the leading coefficient (1). The possible rational roots are therefore \(\pm 1, \pm 2, \pm 4, \pm 8\).

Step 2: Test Rational Roots

We need to test each possible rational root from Step 1 by substituting into the polynomial \(x^{3}-x^{2}-10x-8\). By substituting \(x = -1\), we find \((-1)^3 - (-1)^2 - 10(-1) - 8 = 0\). Thus, \(x = -1\) is a root.

Step 3: Perform Polynomial Division

Now, we will divide the original polynomial \(x^{3}-x^{2}-10x-8\) by \(x + 1\) using synthetic division. The quotient is \(x^2 - 2x - 8\).

Step 4: Factor the Quadratic Equation

Factor the quadratic equation \(x^2 - 2x - 8\). This can be factored into \((x - 4)(x + 2)\).

Step 5: Find Remaining Roots

Set each factor obtained in Step 4 equal to zero and solve: \(x - 4 = 0\) gives \(x = 4\), and \(x + 2 = 0\) gives \(x = -2\). Thus, the roots are \(x = -1, x = 4, x = -2\).

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a powerful tool used in algebra to identify the potential rational solutions to a polynomial equation. This theorem helps in narrowing down the list of possible real roots, making it easier to find exact solutions.
To understand how the Rational Root Theorem works, consider that any rational solution to a polynomial equation with integer coefficients can be expressed as \( \frac{p}{q} \). Here, \( p \) is a factor of the constant term, and \( q \) is a factor of the leading coefficient.
For example, in the polynomial equation \( x^3 - x^2 - 10x - 8 = 0 \), the constant term is \(-8\) and the leading coefficient is \(1\). This means that possible rational roots are the factors of \(-8\) divided by the factors of \(1\), which are:

• \(\pm 1\)
• \(\pm 2\)
• \(\pm 4\)
• \(\pm 8\)

Using the theorem, you can then test these possible values to see if they are indeed roots of the polynomial. This method effectively reduces guesswork and allows for a more systematic approach to finding roots.

Synthetic Division

Once a potential rational root has been identified, synthetic division provides a streamlined method to divide the polynomial by a binomial of the form \(x - c\), where \(c\) is a root. Synthetic division is often preferred over long division because it is quicker and less cumbersome.
Let's say we discovered \(x = -1\) as a root for our polynomial \(x^3 - x^2 - 10x - 8\). We would use synthetic division to divide the cubic polynomial by \(x + 1\). The steps involve:

• Setting up a row with the coefficients of the polynomial: \(1, -1, -10, -8\).
• Using \(-1\) as our divisor.
• Performing arithmetic operations to find the resulting coefficients of the quotient.

This process reveals that the quotient polynomial is \(x^2 - 2x - 8\). Synthetic division not only confirms that \(x = -1\) is a root but also helps in breaking down the polynomial into simpler factors, making it easier to find the remaining roots.

Factoring Quadratics

After using synthetic division, we often end up with a simpler polynomial to solve, such as a quadratic equation. Quadratic equations can typically be solved by factoring them into products of simpler expressions.
Given the quotient from our synthetic division, \(x^2 - 2x - 8\), the goal is to express this quadratic as the product of two binomials. We look for two numbers that multiply to \(-8\) and add to \(-2\), which are \(-4\) and \(2\). Therefore, the quadratic can be factored as \((x - 4)(x + 2)\).
Finally, setting each factor equal to zero gives the solutions \(x = 4\) and \(x = -2\). Factoring quadratics effectively breaks down a problem into smaller, more manageable parts, allowing us to solve for any remaining roots efficiently. This method clarifies the process of reaching a complete solution for polynomial equations.