Problem 32 - Precalculus: Functions and Graphs

Question

Use Descartes' rule of signs to determine the number of possible positive, negative, and non real complex solutions of the equation. $$2 x^{4}-x^{3}+x^{2}-3 x+4=0$$

Step-by-Step Solution

Answer

4, 2, or 0 positive, 0 negative, 0, 2, or 4 complex solutions.

Step 1: Identify the Polynomial

The given polynomial equation is \( f(x) = 2x^4 - x^3 + x^2 - 3x + 4 \). We need to use Descartes' rule of signs to determine the number of real and complex roots.

Step 2: Determine Sign Changes in Original Polynomial

Write down the signs of each term in \( f(x) \): \(+, -, +, -, +\). Count the number of sign changes between consecutive terms. Here they are: 1. From \(+\) to \(-\) (1st change)2. From \(-\) to \(+\) (2nd change)3. From \(+\) to \(-\) (3rd change)4. From \(-\) to \(+\) (4th change)Therefore, there are 4 sign changes.

Step 3: Descartes' Rule for Positive Roots

According to Descartes' Rule of Signs, the number of positive real roots is equal to the number of sign changes or less than it by an even number. Thus, the number of possible positive roots could be 4, 2, or 0.

Step 4: Substitute for Negative Roots

To find the number of negative roots, substitute \(-x\) into the polynomial to get \( f(-x) = 2(-x)^4 - (-x)^3 + (-x)^2 - 3(-x) + 4 \), which simplifies to \( 2x^4 + x^3 + x^2 + 3x + 4 \).

Step 5: Determine Sign Changes for Negative Roots

Write down the signs of each term in \( f(-x) \): \(+,+,+,+,+\). Count the number of sign changes. Here, there are no sign changes, indicating 0 possible negative roots.

Step 6: Determine Possible Non-real Complex Roots

The polynomial is a quartic (degree 4), hence it has 4 roots in total. From the possible positive and negative roots (4, 2, or 0 positive roots and 0 negative roots), the rest of the roots must be non-real complex. If there are 4 positive roots, there are 0 complex roots; if there are 2 positive roots, there are 2 complex roots; if there are 0 positive roots, there are 4 complex roots.

Key Concepts

Polynomial Roots

When dealing with polynomial equations, finding the roots, or solutions, is a fundamental concept. Roots of a polynomial are the values of the variable that satisfy the equation, such that the polynomial equals zero. For example, in the equation \(2x^4 - x^3 + x^2 - 3x + 4 = 0\), the roots are the values of \(x\) for which the polynomial evaluates to zero.

Polynomials can have multiple roots, depending on their degree. A polynomial of degree \(n\) can have up to \(n\) roots. These roots can be real or complex, and Descartes' rule of signs helps in predicting the number of positive and negative real roots. Understanding the nature of these roots is essential for solving polynomials and analyzing their behavior.

Positive Real Roots

Positive real roots of a polynomial are those that are greater than zero and can be found using Descartes' rule of signs. This rule involves analyzing the sign changes in the sequence of coefficients of the polynomial.

For the polynomial \( f(x) = 2x^4 - x^3 + x^2 - 3x + 4 \), the sequence of signs is \(+,-,+,-,+\). Descartes' rule states that the number of positive real roots is equal to the number of sign changes or less by an even number. Here, there are 4 sign changes, which means there could be 4, 2, or 0 positive real roots. It’s important to note that we don’t identify which roots are positive, only the potential number based on sign changes.

Negative Roots

Negative roots of a polynomial can be found by substituting \(-x\) into the polynomial and analyzing the resulting sign changes. This indicates the number of negative or zero roots.

For the polynomial \( f(x) = 2x^4 - x^3 + x^2 - 3x + 4 \), substituting \(-x\) gives \(f(-x) = 2x^4 + x^3 + x^2 + 3x + 4\). The signs of this new sequence are \(+,+,+,+,+\). Since there are no sign changes in \(f(-x)\), it implies that there are 0 negative roots. This tells us that none of the roots of the original polynomial are less than zero.

Complex Roots

Complex roots come into play when the polynomial roots are not real. A polynomial of degree \(n\) will have exactly \(n\) roots, counting multiplicities and complex numbers. If the real roots (both positive and negative) do not account for all \(n\) roots, the remainder must be complex.

For our quartic polynomial \(2x^4 - x^3 + x^2 - 3x + 4\), we have determined possible positive and negative real roots, dismissing any negative roots. Depending on the number of positive real roots (as predicted by Descartes' rule: 4, 2, or 0), the complex roots adjust accordingly:

• If there are 4 positive roots, there would be 0 complex roots
• If there are 2 positive roots, there would be 2 complex roots
• If there are 0 positive roots, all 4 roots would be complex

Complex roots often occur in conjugate pairs, which means they come in pairs like \(a + bi\) and \(a - bi\). Understanding the presence of complex roots is crucial for a complete picture of the solutions to polynomial equations.