Problem 15 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. \(15-24\) : Find all solutions of the equation. $$ x^{3}-x^{2}-10 x-8=0 $$

Step-by-Step Solution

Answer

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

Step 1: Identify Possible Rational Roots

The Rational Root Theorem suggests that possible rational roots are the factors of the constant term (-8) over the factors of the leading coefficient (1). The possible rational roots are \( \pm 1, \pm 2, \pm 4, \pm 8 \).

Step 2: Test Rational Roots

Test these possible roots in the equation to see if any of them is a real root. Using synthetic division or substitution, you find that \( x = -1 \) is a root because it satisfies the equation, making the left-hand side equal to 0.

Step 3: Divide the Polynomial by \( x + 1 \)

Now that \( x = -1 \) is a root, we can perform polynomial division of \( x^3-x^2-10x-8 \) by \( x + 1 \) to find the quotient polynomial. The division results in \( x^2 - 2x - 8 \).

Step 4: Solve the Quadratic Equation

The quotient \( x^2 - 2x - 8 \) is a quadratic equation. We can solve it using the factorization method: \( x^2 - 2x - 8 = (x - 4)(x + 2) \). Thus, the roots are \( x = 4 \) and \( x = -2 \).

Step 5: Combine All Roots

With all the factors found, the complete set of roots of the original cubic equation is \( x = -1, 4, -2 \).

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a handy tool when working with polynomial equations. It gives you a way to find all possible rational solutions to a polynomial equation. The theorem states that any potential rational root of a polynomial equation, like the one given above, is a fraction \( \frac{p}{q} \), where \( p \) is a factor of the constant term and \( q \) is a factor of the leading coefficient.
In the given problem, the constant term is \(-8\) and the leading coefficient is \(1\). This means that the possible rational roots are the factors of \(-8\), which are \( \pm 1, \pm 2, \pm 4, \pm 8 \), since \(1\) is the only factor of itself.

By identifying these potential roots, the theorem narrows down the list of numbers you would need to check, making it easier to determine if a polynomial has rational solutions.

• This step eliminates much of the guesswork.
• Saves time when solving equations.

It is worth noting that while this theorem lists possible rational roots, it doesn't guarantee all of them will work. They are merely candidates to be tested.

Polynomial Division

Once a rational root, such as \( x = -1 \), is found, the next step is using polynomial division. This division helps further simplify the polynomial equation, letting you work with reduced terms.
Polynomial division here is somewhat similar to long division, but instead of dividing numbers, you're dividing terms. It's a way to "divide out" a known factor from the polynomial, like \( x + 1 \) in the given exercise, which is associated with the root \( x = -1 \).

By dividing \( x^3-x^2-10x-8 \) by \( x + 1 \), you get the quotient polynomial \( x^2 - 2x - 8 \). This quotient is a quadratic polynomial, making it simpler to solve next using other methods for solving polynomials.

• The process reduces the polynomial degree.
• Leaves you with more manageable terms.

Using polynomial division provides a more streamlined approach to breaking down complex polynomial equations. It sets up the final stages of finding the equation’s complete set of roots.

Quadratic Equations

After polynomial division provides a quadratic equation, like \( x^2 - 2x - 8 \), you're ready to solve it. Quadratic equations are fundamental and can be solved by various methods such as factoring, completing the square, or using the quadratic formula.
For this exercise, the solution involves factoring. Factoring is a process where you express the quadratic as a product of its roots or factors. By setting the equation \( x^2 - 2x - 8 \) equal to zero and looking for two numbers that multiply to \(-8\) and add up to \(-2\), you find that the equation can be factored as:

• \( (x - 4)(x + 2) = 0 \)

Solving these factors gives the roots \( x = 4 \) and \( x = -2 \).

Quadratic equations like this are often easier and faster to solve compared to higher-degree polynomials.

• When factoring is possible, it simplifies finding solutions.
• Shows the roots directly through simple multiplication reversal.

Thus, factoring provides a direct way to determine the extra roots for a polynomial that has already been reduced to a quadratic form.