Problem 27 - 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. $$4 x^{3}-6 x^{2}+x-3=0$$

Step-by-Step Solution

Answer

3 or 1 positive roots, 0 negative roots, and up to 2 non-real complex roots.

Step 1: Count Sign Changes for Positive Roots

To determine how many positive real roots the polynomial may have, note the coefficients of the polynomial: \(4x^3 - 6x^2 + x - 3\). Observe the signs of these coefficients: \(+, -, +, -\). Here we have a total of three sign changes: 1. \(+\) to \(-\) from \(4x^3\) to \(-6x^2\)2. \(-\) to \(+\) from \(-6x^2\) to \(+x\)3. \(+\) to \(-\) from \(+x\) to \(-3\)According to Descartes' rule of signs, this implies there could be 3, 1, or 0 positive real roots (3 minus any even number).

Step 2: Change Variable to Determine Negative Roots

To determine the number of negative real roots, we substitute \(x\) with \(-x\) leading to the polynomial \(4(-x)^3 - 6(-x)^2 + (-x) - 3 = -4x^3 - 6x^2 - x - 3\). Simplify it to the form \(-4x^3 - 6x^2 - x - 3\). Now check for sign changes in the coefficients: \(-, -, -, -\), which shows no sign changes.Hence, there are 0 negative real roots possible.

Step 3: Calculate Non-Real Complex Roots

For a cubic polynomial, the total number of roots (including real and complex) is equal to the degree of the polynomial, which is 3. From Step 1, there can be up to 3, 1, or 0 positive real roots and from Step 2, there are no negative real roots. Depending on the real roots, the number of non-real complex roots can be either 2 (if there is 1 positive real root), 0 (if there are 3 positive real roots), or none depending on the actual distribution of roots, but by default assuming maximum potential complex roots when solving theoretically should be considered if there is a single positive root.

Key Concepts

Positive Real Roots

When solving a polynomial equation, such as the cubic equation given, identifying the number of positive real roots is a critical step. Descartes' Rule of Signs assists by analyzing the sign changes in the sequence of the polynomial's coefficients. For the polynomial \(4x^3 - 6x^2 + x - 3\), we examine the sequence:

• Start with the coefficient of \(x^3\), which is \(+4\).
• Then, move to the coefficient of \(x^2\), which is \(-6\).
• Next, the coefficient of \(x\) is \(+1\).
• Finally, the constant term is \(-3\).

The pattern of signs is \(+, -, +, -\), revealing three sign changes:

• From \(+4\) to \(-6\)
• From \(-6\) to \(+1\)
• From \(+1\) to \(-3\)

According to Descartes' Rule of Signs, the number of positive real roots could be 3, 1, or 0, subtracting an even number from the total number of sign changes.

Negative Real Roots

Negative real roots are evaluated by substituting \(x\) with \(-x\) in the polynomial equation. This creates a change in the expression, allowing us to analyze the new sequence of coefficients for sign changes to determine the number of potential negative real roots. After substituting \(x\) with \(-x\) in the equation \(4x^3 - 6x^2 + x - 3\), the resulting polynomial is \(-4x^3 - 6x^2 - x - 3\). Observing this new sequence:

• The coefficient for \(x^3\) is \(-4\).
• The coefficient for \(x^2\) is also \(-6\).
• The coefficient for \(x\) is \(-1\).
• The constant term remains \(-3\).

As we can see, all signs are negative \(-, -, -, -\), which results in zero sign changes. Consequently, using Descartes' Rule of Signs, there are no possible negative real roots for this polynomial equation.

Non-Real Complex Roots

To find the total number of solutions for a polynomial, including non-real complex roots, we look at the degree of the polynomial. For our cubic equation, the degree is 3. This means it has a total of 3 roots, real or complex. Considering the possible real roots determined earlier:

• There could be up to 3, 1, or 0 positive real roots.
• There are confirmed to be 0 negative real roots.

Depending on how many positive real roots exist, the remaining roots will be non-real complex roots. For example:

• If there is 1 positive real root, then there are likely 2 non-real complex roots.
• If there are 3 positive real roots, there would be no complex roots.

In general, a polynomial's non-real complex roots appear in conjugate pairs, maintaining the total count of roots constrained by the polynomial's degree.