Problem 72 - Precalculus Mathematics for Calculus

Question

Show that the given values for \(a\) and \(b\) are lower and upper bounds for the real zeros of the polynomial. $$P(x)=x^{4}-2 x^{3}-9 x^{2}+2 x+8 ; \quad a=-3, b=5$$

Step-by-Step Solution

Answer

-3 is a lower bound, but 5 is not an upper bound.

Step 1: Evaluate Polynomial at Lower Bound

Substitute the lower bound value, \( a = -3 \), into the polynomial \( P(x) \). Calculate \( P(-3) \). Determine whether the result indicates \( a \) is a lower bound by checking if \( P(-3) > 0 \).

Step 2: Calculation of P(-3)

Compute the polynomial \( P(x) = x^4 - 2x^3 - 9x^2 + 2x + 8 \) at \( x = -3 \):\[ P(-3) = (-3)^4 - 2(-3)^3 - 9(-3)^2 + 2(-3) + 8 \]\[ = 81 + 54 - 81 - 6 + 8 \]\[ = 56 \]Since \( P(-3) = 56 > 0 \), \( a = -3 \) is indeed a lower bound.

Step 3: Evaluate Polynomial at Upper Bound

Substitute the upper bound value, \( b = 5 \), into the polynomial \( P(x) \). Calculate \( P(5) \). Determine whether the result indicates \( b \) is an upper bound by checking if \( P(5) < 0 \).

Step 4: Calculation of P(5)

Compute the polynomial \( P(x) = x^4 - 2x^3 - 9x^2 + 2x + 8 \) at \( x = 5 \):\[ P(5) = (5)^4 - 2(5)^3 - 9(5)^2 + 2(5) + 8 \]\[ = 625 - 250 - 225 + 10 + 8 \]\[ = 168 \]Since \( P(5) = 168 > 0 \), \( b = 5 \) suggests \( b = 5 \) is not an upper bound.

Step 5: Conclusion

The evaluation of \( P(x) \) at \( a = -3 \) shows it is a lower bound because \( P(-3) > 0 \). However, \( b = 5 \) is not an upper bound since \( P(5) > 0 \). The upper bound assessment fails.

Key Concepts

Polynomial Evaluation

When we talk about evaluating a polynomial, we're examining the value of a polynomial function at a specific point. This involves substituting a given number for the variable in the polynomial equation and then performing the arithmetic calculations.
In the case of the exercise, the polynomial to evaluate is given as

• P(x) = x^4 - 2x^3 - 9x^2 + 2x + 8

To evaluate this polynomial at a specific value, say at a lower bound of

• a = -3

we substitute

• x = -3

in the polynomial. Then, the goal is to determine whether the polynomial evaluates to a positive or negative number, indicating important characteristics about bounds and roots.
This evaluation helps us understand the behavior of the polynomial on a number line and its implications on real roots.

Real Zeros

The real zeros of a polynomial are the points where the polynomial equals zero when evaluated. These are essentially the x-values where the graph of the polynomial touches or crosses the x-axis. Finding real zeros is a critical aspect in polynomial analysis because they help in analyzing the polynomial's roots and its overall graph.
In practical terms, real zeros can give insights into the solutions of polynomial equations. They can be determined or estimated using various methods, such as factoring, synthetic division, or graphical analysis.
Understanding real zeros assists in narrowing down the intervals where the bound of zeros lies, and with proper evaluation, can confirm if a certain value is a bound for any real zero or not.

Algebraic Bounds

Algebraic bounds are essentially estimated limits within which the real zeros of a polynomial are found. In our exercise, algebraic bounds are determined via specific test values, like

• a = -3
• b = 5

These bounds help indicate the range within which the real zeros exist. To confirm these bounds, we substitute these boundary values into the polynomial.
If the polynomial value is greater than zero at the lower bound, this suggests it might be a true lower bound, which is confirmed in our example with

• P(-3) > 0

However, for

• b = 5

where

• P(5) > 0

the evaluation indicates that this is not a valid upper bound. Accurate algebraic bounds are essential because they simplify finding the real zeros by reducing the range under consideration.

Polynomial Roots

The roots of a polynomial are the x-values for which the polynomial equals zero. In simpler terms, these are the solutions of the polynomial equation. Polynomial roots can be real or complex, but in the practical context of solving equations, we often focus on real roots.
Roots are crucial as they provide the exact points where the polynomial graph intersects the x-axis. They are determined by setting the polynomial equal to zero and solving for x, either algebraically or using numerical methods.
Furthermore, assessing polynomial roots involves applying algebraic rules such as the Descartes' Rule of Signs or the Intermediate Value Theorem to understand the number and location of real roots. This is particularly vital when establishing bounds and evaluating the performance of a polynomial under certain test points.
In this exercise, the attempt was to check bounds, but accurately determining or confirming roots goes far beyond just checking a few values, requiring deeper analysis or tools.