Problem 23 - A Graphical Approach to Precalculus with Limits

Question

Solve each equation analyrically for all complex solutions, giving exact forms in your solution set. Then graph the left side of the equation as \(Y_{1}\) in the suggested viewing window and, using the capabilities of vour calculater, suppert the real solutions. $$\begin{aligned}&x^{3}-x^{2}-64 x+64=0\\\&[-10,10] \text { by }[-300,300]\end{aligned}$$

Step-by-Step Solution

Answer

Roots are \( x = 1, 8, -8 \). Graph confirms these roots.

Step 1: Identify the Polynomial

The given equation is a cubic polynomial: \( x^3 - x^2 - 64x + 64 = 0 \). We need to find the roots of this cubic equation.

Step 2: Check for Rational Roots

Using the Rational Root Theorem, possible rational roots are the factors of 64, i.e., \( \pm 1, \pm 2, \pm 4, \pm 8, \pm 16, \pm 32, \pm 64 \). Check these values in the polynomial to find a root.

Step 3: Verify Rational Root

By substituting potential roots into the equation: \( x = 1 \) works, giving \( 1^3 - 1^2 - 64 \times 1 + 64 = 0 \). So, \( x = 1 \) is a root.

Step 4: Perform Synthetic Division

Using \( x = 1 \) as a root, perform synthetic division on \( x^3 - x^2 - 64x + 64 \) by \( x - 1 \), resulting in \( x^2 - 64 \).

Step 5: Solve the Quotient Quadratic Equation

The quotient from the division is \( x^2 - 64 = 0 \). This simplifies to \( x^2 = 64 \). Find \( x \) by taking the square root: \( x = \pm 8 \). Hence, the roots are \( x = 8 \) and \( x = -8 \).

Step 6: Combine and State All Roots

Combine the roots from previous steps: the complete set of solutions to the polynomial is \( x = 1, 8, -8 \).

Step 7: Graph the Polynomial

Graph \( x^3 - x^2 - 64x + 64 \) on the calculator in the suggested window [-10, 10] by [-300, 300]. The graph should cross the x-axis at the real roots \( x = 1, 8, -8 \).

Step 8: Verify Real Solutions Using Calculator

Using the graphing capabilities of the calculator, check that the curve crosses the x-axis at \( x = 1, 8, -8 \), confirming the real solutions found analytically.

Key Concepts

Polynomial Roots

Every polynomial, like our cubic equation, seeks its own roots. But what exactly are polynomial roots? These are the values for which a polynomial (in this case, a cubic polynomial) equals zero.
When you solve a polynomial equation, you're essentially finding where its graph crosses the x-axis. These points, known as roots or zeros, are where the polynomial's value is zero.

To find roots, we often face the challenge of the degree of the polynomial:

• Linear equations (degree 1) have exactly one root.
• Quadratic equations (degree 2) have up to two roots.
• Cubic equations (like our example) can have up to three roots.

The degree of the polynomial also tells us the maximum number of roots to expect. Furthermore, not all roots are real numbers; some may be complex, especially when dealing with polynomials of higher degree.

Rational Root Theorem

The Rational Root Theorem is a powerful tool that tells us which rational numbers could potentially be roots of a polynomial. This theorem states that any rational root of a polynomial equation with integer coefficients will be a factor of the constant term divided by a factor of the leading coefficient.

For the cubic polynomial equation we are dealing with, the constant term is 64, and the leading coefficient is 1. Thus, according to the Rational Root Theorem, the candidates for rational roots are

• All the factors of 64 (like \(\pm 1, \pm 2, \pm 4, \pm 8, \pm 16, \pm 32, \pm 64\)).

Candidates are tested by substituting them into the polynomial. If the substitution results in zero, then it confirms that the candidate is a root. In this case, through substitution, it was verified that 1 is indeed a root of the polynomial.

Synthetic Division

Once we've identified a root, synthetic division helps to simplify the polynomial. It's a quicker alternative to long division for polynomials. By using this method, we can divide the polynomial by \(x - r\) where \(x = r\) is a known root.

Here's how it works in our problem:

• First, take the root found (\(x = 1\)) and divide the polynomial \(x^3 - x^2 - 64x + 64\) by \(x - 1\).
• Synthetic division yields the quotient \(x^2 - 64\).

This process reduces the original polynomial degree from three to two, making it easier to find remaining roots. You align coefficients, drop numbers, and multiply back, simplifying as you go.

Quadratic Equations

After synthetic division, we often encounter a quadratic polynomial, which is simpler to solve. A quadratic equation, like the resultant \(x^2 - 64\), usually leads us to its roots by setting it to zero: \(x^2 - 64 = 0\).

To solve this quadratic equation:

• Move the constant to the other side: \(x^2 = 64\).
• Take the square root of both sides to solve for \(x\): \(x = \pm 8\).

Quadratic equations can be solved using methods such as factoring, completing the square, or applying the quadratic formula. For simple cases like ours, finding the square root is direct and effective.
This solution reveals the two additional roots of the original cubic polynomial: \(x = 8\) and \(x = -8\).