Problem 56 - Algebra for College Students

Question

Use the rational root theorem, Descartes' rule of signs, and the theorem on bounds as aids in finding all solutions to each equation. $$x^{3}-7 x^{2}+17 x-15=0$$

Step-by-Step Solution

Answer

The solutions are \( x = 3 \), \( x = 2 + i \), and \( x = 2 - i \).

Step 1 - Identify Potential Rational Roots

According to the Rational Root Theorem, potential rational roots are the factors of the constant term (15) divided by the factors of the leading coefficient (1). The possible rational roots are ±1, ±3, ±5, ±15.

Step 2 - Use Descartes' Rule of Signs for Positive Roots

Descartes' Rule of Signs indicates that the number of positive real roots is given by the number of sign changes in the polynomial. For the polynomial \( x^{3} - 7x^{2} + 17x - 15 = 0 \), the signs change twice (from \( +x^3 \) to \( -7x^2 \) and from \( +17x \) to \( -15 \)), suggesting 2 or 0 positive roots.

Step 3 - Use Descartes' Rule of Signs for Negative Roots

To determine the number of negative real roots, evaluate the polynomial at \( -x \). The polynomial becomes \( -x^{3} - 7x^{2} - 17x - 15 \), which has one sign change, indicating exactly 1 negative root.

Step 4 - Test Potential Roots

Test the potential roots from step 1 using synthetic division or direct substitution. For example, substitute \( x = 1 \): \( 1^3 - 7(1)^2 + 17(1) - 15 = -4 \). Thus, \( x = 1 \) is not a root. Repeat this process for other potential roots.

Step 5 - Find Actual Roots

By testing, we find \( x = 3 \) is an actual root as \( 3^3 - 7(3)^2 + 17(3) - 15 = 0 \). Perform synthetic division to factor the polynomial: \( x^3 - 7x^2 + 17x - 15 = (x - 3)(x^2 - 4x + 5) \).

Step 6 - Solve the Quadratic Equation

Solve the quadratic equation \( x^2 - 4x + 5 = 0 \) using the quadratic formula: \[ x = \frac{4 ± \sqrt{4^2 - 4(1)(5)}}{2(1)} \] \[ x = \frac{4 ± \sqrt{16 - 20}}{2} \] \[ x = \frac{4 ± \sqrt{-4}}{2} \] \[ x = 2 ± i \].

Step 7 - Compile All Solutions

Combine the solutions found. The solutions to the equation \( x^3 - 7x^2 + 17x - 15 = 0 \) are \( x = 3 \), \( x = 2 + i \), and \( x = 2 - i \).

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a handy tool for finding potential rational roots of a polynomial equation. It tells us that any potential rational root, written as \( \frac{p}{q} \), must have a numerator \( p \) that is a factor of the constant term and a denominator \( q \) that is a factor of the leading coefficient.

In our example, the polynomial equation is \( x^3 - 7x^2 + 17x - 15 = 0 \). The constant term here is 15, and the leading coefficient is 1.

• Factors of 15: ±1, ±3, ±5, ±15
• Factors of 1: ±1

The possible rational roots are obtained by dividing the factors of the constant term by the factors of the leading coefficient. So, we get: ±1, ±3, ±5, ±15.

These are the potential rational roots we need to test to see if they are actual roots of the polynomial.

Descartes' Rule of Signs

Descartes' Rule of Signs helps us determine the number of positive and negative real roots a polynomial can have. It looks at the number of sign changes in the polynomial's coefficients.

For positive roots, we check the original polynomial: \( x^3 - 7x^2 + 17x - 15 = 0 \).

• Sign changes: Positive to negative, then back to positive, and then to negative again. So, there are 2 sign changes.

This indicates there could be 2 or 0 positive real roots.

For negative roots, we plug in \( -x \) and check the polynomial: \( -x^3 - 7x^2 - 17x - 15 = 0 \).

• Sign changes: There is only 1 sign change.

This tells us there is exactly 1 negative real root.

Synthetic Division

Synthetic division is used to test potential roots by dividing the polynomial by \( x - r \), where \( r \) is the root being tested. It's a simplified manual method to divide a polynomial.

To test a root, for instance, \( x = 3 \):

• Write down the coefficients of the polynomial: 1, -7, 17, -15.
• Write 3 outside the division box.
• Bring down the leading coefficient (1) unchanged.
• Multiply 3 by the previously found number and put the result under the next coefficient.
• Add this result to the next coefficient.
• Repeat the process for all coefficients.

If the final result (the remainder) is 0, then \( x = 3 \) is a root.

Quadratic Formula

The quadratic formula helps solve quadratic equations of the form \( ax^2 + bx + c = 0 \). The formula is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In our example, after finding a root using synthetic division, we are left with the quadratic equation:
\( x^2 - 4x + 5 = 0 \).

Here, \( a = 1 \), \( b = -4 \), and \( c = 5 \). Plugging these values into the quadratic formula gives:
\[ x = \frac{4 \pm \sqrt{(4)^2 - 4(1)(5)}}{2(1)} \]
\[ x = \frac{4 \pm \sqrt{16 - 20}}{2} \]
\[ x = \frac{4 \pm \sqrt{-4}}{2} \]
\[ x = 2 \pm i \].
These are complex roots.

Complex Numbers

Complex numbers include both a real part and an imaginary part and are written in the form \( a + bi \), where \( i \) is the imaginary unit (\( i^2 = -1 \)).

In our solution, we find the roots \( x = 2 + i \) and \( x = 2 - i \). These are complex because they include the imaginary unit \( i \).

Imaginary numbers arise from the square root of negative numbers. In the quadratic formula, when the term under the square root (discriminant) is negative, the solution involves imaginary numbers.

Combining these complex roots with the real root, the polynomial \( x^3 - 7x^2 + 17x - 15 = 0 \) has three solutions:

• One real root: \( x = 3 \)
• Two complex roots: \( x = 2 + i \) and \( x = 2 - i \).

Understanding complex numbers is crucial in solving higher-degree polynomials with no real solutions.