Problem 48 - College Algebra with Corequisite Support

Question

For the following exercises, use Descartes' Rule to determine the possible number of positive and negative solutions. Confirm with the given graph. $$ f(x)=x^{3}-2 x^{2}-5 x+6 $$

Step-by-Step Solution

Answer

There can be 2 or 0 positive roots and 1 negative root.

Step 1: Determine Sign Changes for Positive Roots

To find the number of possible positive solutions, examine how the signs of the coefficients of the polynomial \( f(x) = x^3 - 2x^2 - 5x + 6 \) change:1. The coefficient sequence in terms of signs is: \( +, -, -, + \).2. There are 2 sign changes here (from \( + \) to \( - \) and from \( - \) to \( + \)).3. Hence, there can be 2 or 0 positive real roots.

Step 2: Determine Sign Changes for Negative Roots

To find the number of negative solutions, substitute \( x = -y \) into the polynomial to find \( f(-y) \):\[f(-y) = (-y)^3 - 2(-y)^2 - 5(-y) + 6 = -y^3 - 2y^2 + 5y + 6\]Examine the sign changes:1. The coefficient sequence for \( f(-y) \) is: \(-, -, +, + \).2. There is 1 sign change (from \( - \) to \( + \)).3. Hence, there can be 1 negative real root.

Step 3: Use Graphical Confirmation

Upon analyzing the graph of \( f(x) \), observe the points where the graph crosses the x-axis. This corresponds to the real roots of the polynomial.

Key Concepts

Polynomial Roots

Polynomial roots are the values of **x** where a polynomial equation equals zero. For instance, for the polynomial \(f(x) = x^3 - 2x^2 - 5x + 6\), the roots are where \(f(x) = 0\). These roots are crucial because they tell us the solutions or **x**-intercepts of the polynomial. To visualize, if you were to graph this polynomial function, the roots are the points where the graph intersects the x-axis.
Identifying the number and type (real or complex) of these roots can often be predicted by Descartes' Rule of Signs. While complex roots occur in conjugate pairs, real roots can either be positive, negative, or zero. Each root is a solution to the equation and can be thought of as where the function "touches" or "crosses" the x-axis. Knowing this helps in understanding the overall behavior of the polynomial graph.

Positive Solutions

Positive solutions for a polynomial are those roots that are greater than zero. In other words, these roots are found somewhere to the right of the origin on the x-axis of a graph. Descartes' Rule of Signs gives us a systematic way to estimate how many positive real roots a polynomial could have.
As in the example \(f(x) = x^3 - 2x^2 - 5x + 6\), we examine the sign changes of the coefficients. Here, the sequence **+, -, -, +** shows two changes, implying there could be either 2 or 0 positive solutions. The concept here is fairly intuitive: if no sign change occurs, no switch from positive to negative or vice versa in the function occurs as it crosses the x-axis, indicating no positive (or negative) roots.

• You might have precisely the number of sign changes as positive solutions.
• Alternatively, the actual number of positive solutions could be fewer by an even number.

Negative Solutions

Negative solutions, unlike positive ones, are roots that lie to the left of the origin on the graph. For these, if you measure solutions on the x-axis, they exist where the graph falls below zero. To find the potential negative solutions using Descartes' Rule of Signs, you substitute \(x = -y\) into the polynomial, effectively flipping the graph around the y-axis.
Taking our example, substituting leads to \(f(-y) = -y^3 - 2y^2 + 5y + 6\), with the sign sequence **-, -, +, +**. Here, only one sign change occurs, suggesting a single negative solution. Again, remember the rule about number reduction: the actual number of negative solutions might be fewer than the number of sign changes, reduced by an even number.

• For negative solutions, examining \(-y\) is a quick substitution method provided by Descartes' Rule.
• Sign changes directly relate to potential shifts on the graph across the y-axis, showing potential negative root locations.