Finding Zeros: A Polynomial Exploration

Nov 8, 2025 by ADMIN 40 views

Let's dive into the exciting world of polynomial functions! In this article, we're going to tackle the challenge of finding the zeros and their multiplicities for the polynomial function f(x) = 2x⁵ - 11x⁴ + x³ + 62x² - 78x + 24. We'll also explore how to use Descartes' Rule of Signs and the Upper and Lower Bound Theorem to make our search for rational zeros a whole lot easier. So, buckle up, and let's get started!

Understanding the Problem

Before we start crunching numbers, let's make sure we understand what we're trying to achieve. Zeros of a polynomial function are the values of x that make the function equal to zero, i.e., f(x) = 0. These zeros are also the x-intercepts of the graph of the polynomial. The multiplicity of a zero tells us how many times that zero appears as a root of the polynomial. This affects the behavior of the graph near that x-intercept. For example, if a zero has a multiplicity of 1, the graph crosses the x-axis at that point. If it has a multiplicity of 2, the graph touches the x-axis and turns around. Now that we know what the key terms mean, let's get started on the mathematical method.

Descartes' Rule of Signs

Descartes' Rule of Signs is a handy tool that helps us predict the number of positive and negative real zeros of a polynomial. It's like having a sneak peek at the possible solutions! Here's how it works:

Positive Real Zeros: Count the number of sign changes in the coefficients of f(x). The number of positive real zeros is either equal to this count or less than this count by an even number. For f(x) = 2x⁵ - 11x⁴ + x³ + 62x² - 78x + 24, the signs are +, -, +, +, -, +. There are 4 sign changes, so there are either 4, 2, or 0 positive real zeros.
Negative Real Zeros: Count the number of sign changes in the coefficients of f(-x). The number of negative real zeros is either equal to this count or less than this count by an even number. Let's find f(-x): f(-x) = 2(-x)⁵ - 11(-x)⁴ + (-x)³ + 62(-x)² - 78(-x) + 24 = -2x⁵ - 11x⁴ - x³ + 62x² + 78x + 24. The signs are -, -, -, +, +, +. There is 1 sign change, so there is exactly 1 negative real zero.

So, based on Descartes' Rule of Signs, we expect to find either 4, 2, or 0 positive real zeros and exactly 1 negative real zero. This narrows down our search quite a bit! Keep in mind that this rule only gives us possibilities, not guarantees. Also, it doesn't tell us anything about non-real complex roots. Knowing how many roots the polynomial might have helps us to look for roots in the correct area.

Upper and Lower Bound Theorem

The Upper and Lower Bound Theorem helps us find boundaries for the real zeros of a polynomial. It tells us that if we divide a polynomial by (x - c), we can determine whether c is an upper or lower bound for the real zeros. Here's the gist:

Upper Bound: If we divide f(x) by (x - c), where c > 0, using synthetic division, and all the numbers in the last row (the quotient and the remainder) are either positive or zero, then c is an upper bound for the real zeros. This means there are no real zeros greater than c.
Lower Bound: If we divide f(x) by (x - c), where c < 0, using synthetic division, and the numbers in the last row alternate in sign (with zero considered either positive or negative), then c is a lower bound for the real zeros. This means there are no real zeros less than c.

This theorem is super useful because it can prevent us from wasting time testing values that are outside the range of possible real zeros. We can efficiently narrow our search to the values that are actually possible.

Finding Rational Zeros

Now that we have some tools to guide us, let's find the rational zeros of f(x). The Rational Root Theorem tells us that if f(x) has rational zeros, they must be of the form p/q, where p is a factor of the constant term (24) and q is a factor of the leading coefficient (2). So, the possible rational zeros are:

±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24, ±1/2, ±3/2

Let's start testing these values using synthetic division. We'll also keep in mind Descartes' Rule of Signs and the Upper and Lower Bound Theorem to make our search more efficient. Let's try x = 1:

2  -11   1   62  -78   24
    2   -9  -8   54  -24
--------------------------
2  -9  -8   54  -24    0

Since the remainder is 0, x = 1 is a zero of f(x). Now we have f(x) = (x - 1)(2x⁴ - 9x³ - 8x² + 54x - 24). Let's continue testing the possible rational zeros on the quotient 2x⁴ - 9x³ - 8x² + 54x - 24. Let's try x = 2:

2  -9  -8   54  -24
    4 -10 -36   36
------------------
2  -5 -18   18   12

Since the remainder is not 0, x = 2 is not a zero. Let's try x = 3:

2   -9   -8   54  -24
    6   -9  -51    9
---------------------
2   -3  -17    3  -15

Since the remainder is not 0, x = 3 is not a zero. Let's try x = 1/2:

2  -9   -8   54  -24
    1   -4   -6   24
---------------------
2  -8  -12   48    0

Since the remainder is 0, x = 1/2 is a zero. Now we have f(x) = (x - 1)(x - 1/2)(2x³ - 8x² - 12x + 48). Factoring out a 2 from the cubic factor, we get f(x) = 2(x - 1)(x - 1/2)(x³ - 4x² - 6x + 24). Let's try x = 4 on the cubic factor:

1   -4   -6   24
    4    0  -24
----------------
1    0   -6    0

Since the remainder is 0, x = 4 is a zero. Now we have f(x) = 2(x - 1)(x - 1/2)(x - 4)(x² - 6). Setting the quadratic factor to zero, x² - 6 = 0, we find x = ±√6. Thus, the zeros are x = 1, 1/2, 4, √6, -√6.

Determining Multiplicities

Now that we've found the zeros, let's determine their multiplicities. By looking at the factored form of the polynomial, we can see that each zero appears only once:

x = 1: Multiplicity 1
x = 1/2: Multiplicity 1
x = 4: Multiplicity 1
x = √6: Multiplicity 1
x = -√6: Multiplicity 1

Therefore, all the zeros have a multiplicity of 1. In other words, the graph of the polynomial crosses the x-axis at all the x-intercepts.

Conclusion

Finding the zeros of a polynomial can be a challenging but rewarding task. By using tools like Descartes' Rule of Signs, the Upper and Lower Bound Theorem, and the Rational Root Theorem, we can efficiently narrow down our search and find all the zeros and their multiplicities. In this case, the zeros of f(x) = 2x⁵ - 11x⁴ + x³ + 62x² - 78x + 24 are x = 1, 1/2, 4, √6, -√6, and each has a multiplicity of 1. Keep practicing, and you'll become a pro at finding zeros in no time!

By following these steps, you can systematically find the zeros and their multiplicities for any polynomial function. Remember to use all the tools at your disposal and to be patient and persistent in your search. Good luck, and have fun exploring the world of polynomials!