Problem 20 - Precalculus

Question

Find the rational roots of each equation, and then solve the equation. (Use the rational roots theorem and the upper and lower bound theorem, as in Example 2.) $$3 x^{3}-16 x^{2}+17 x-4=0$$

Step-by-Step Solution

Answer

The rational roots are \(x = \frac{1}{3}, \frac{2}{3}, 6\).

Step 1: Identify Potential Rational Roots

Using the Rational Root Theorem, identify potential rational roots of the polynomial equation. The theorem states that any rational root of the polynomial equation \(ax^n + bx^{n-1} + ... + k = 0\) is of the form \(\pm \frac{p}{q}\), where \(p\) is a factor of the constant term \(k\) and \(q\) is a factor of the leading coefficient \(a\). For the polynomial \(3x^3 - 16x^2 + 17x - 4 = 0\), \(p\) are the factors of \(-4\) and \(q\) are the factors of \(3\). This gives potential roots: \(\pm 1, \pm 2, \pm 4, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{4}{3}\).

Step 2: Test Potential Roots Using Substitution

Substitute each potential rational root into the polynomial to see if it produces zero. When substituting \(x = 1\), the polynomial \(3(1)^3 - 16(1)^2 + 17(1) - 4 = 0\) does not satisfy. Testing \(x = 2\), we get \(3(2)^3 - 16(2)^2 + 17(2) - 4 = 0\), which is also not zero. Continue testing until a root is found.

Step 3: Confirm Root and Factor Polynomial

After testing, you find that \(x = \frac{1}{3}\) is a root because substituting it gives zero. Use synthetic division or polynomial division to divide the original polynomial by \(x - \frac{1}{3}\) to simplify it into a quadratic, \(3x^2 - 16x + 12\).

Step 4: Solve the Quadratic Equation

Solve the quadratic equation \(3x^2 - 16x + 12 = 0\) by either factoring, using the quadratic formula, or completing the square. Factoring gives \((3x - 2)(x - 6) = 0\). Thus the roots are \(x = \frac{2}{3}\) and \(x = 6\).

Step 5: Compile the Roots

The rational roots of the original polynomial equation \(3x^3 - 16x^2 + 17x - 4 = 0\) are \(x = \frac{1}{3}, \frac{2}{3}, 6\).

Key Concepts

Upper and Lower Bound Theorem

The Upper and Lower Bound Theorem is a useful tool in helping us understand the range within which all real roots of a polynomial can be found. Generally, this theorem gives us a sense of the interval we should be working with when hunting for roots. If we know a specific value does not yield trivial outcomes like zero when checked against a polynomial, it could serve as a boundary marker. Here’s how it works practically: - **Upper Bound**: If we divide the polynomial by \(x - c\) and all the resulting numbers in the synthetic division sequence are non-negative (all zeros included), \(c\) is considered an upper bound for the roots. - **Lower Bound**: Conversely, if each number in the synthetic division sequence alternates sign, then \(c\) is a lower bound.By understanding the bounds, we cut down the number of potential candidates when searching for rational roots using the Rational Root Theorem, greatly optimizing our process.

Synthetic Division

Synthetic division is a simpler and faster alternative to traditional long division for dividing polynomials, particularly handy when we're testing potential roots. It's especially useful with linear divisors like \(x - c\). This method helps us determine if a particular value is a root of the polynomial and can be used to simplify polynomials by dividing out roots. The process involves a few steps: - List the coefficients of the polynomial. - Use the potential root \(c\) in synthetic division: bring down the leading coefficient, multiply it by \(c\), and adjust each subsequent coefficient accordingly.- If the final result (remainder) is zero, then \(c\) is indeed a root.Synthetic division not only helps to confirm a rational root but provides a reduced polynomial, which can further be analyzed or decomposed into simpler factors.

Polynomial Equations

Polynomial equations take the form \(a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 = 0\). Solving these equations involves finding the values of \(x\) (roots) that make the polynomial equal to zero. Here’s how you approach solving a polynomial equation: - **Identify possible rational roots** by using the Rational Root Theorem, which narrows down the possibilities to factors of the constant and leading coefficient. - **Test these potential roots** using substitution or synthetic division to verify which ones satisfy the equation.- **Simplify the polynomial** using techniques like synthetic division to break it down when a root is found.- **Solve the resulting equations**: After reducing the polynomial, solve the simpler quadratic or linear equations.As evidenced by the original example, polynomial equations sometimes require finding multiple feasible values, each of which corresponds to a different root.