Problem 3
Question
State the possible number of positive real zeros, negative real zeros, and imaginary zeros of each function. \(f(x)=5 x^{3}+8 x^{2}-4 x+3\)
Step-by-Step Solution
Verified Answer
2 or 0 positive real zeros, 1 or 0 negative real zero, 0 or 2 imaginary zeros.
1Step 1: Understanding the Function
The given function is a polynomial of degree 3: \(f(x) = 5x^3 + 8x^2 - 4x + 3\). A polynomial of degree \(n\) can have at most \(n\) real zeros and \(n\) complex zeros.
2Step 2: Determine Possible Positive Real Zeros using Descartes' Rule of Signs
Count the sign changes in \(f(x)\) to determine the possible number of positive real zeros. The terms \(5x^3, 8x^2, -4x, 3\) have signs \(+, +, -, +\), respectively. There are two sign changes (from \(+\) to \(-\) and \(-\) to \(+\)). Thus, there can be 2 or 0 positive real zeros.
3Step 3: Determine Possible Negative Real Zeros using Descartes' Rule of Signs
Evaluate \(f(-x)\): \(f(-x) = 5(-x)^3 + 8(-x)^2 - 4(-x) + 3 = -5x^3 + 8x^2 + 4x + 3\). The terms \(-5x^3, 8x^2, 4x, 3\) have signs \(-, +, +, +\). There is one sign change (from \(-\) to \(+\)). Thus, there can be 1 or 0 negative real zeros.
4Step 4: Calculate Possible Imaginary Zeros
The total number of zeros, real or complex, must equal the degree of the polynomial, which is 3. Considering previous possible real zeros (2 or 0 positive and 1 or 0 negative), calculate imaginary zeros: If there are 2 positive and 1 negative real zeros, then there are 0 imaginary zeros. If there are 0 positive and 1 negative real zero, then there are 2 imaginary zeros.
Key Concepts
Polynomial DegreeReal ZerosComplex ZerosImaginary Zeros
Polynomial Degree
Polynomials are mathematical expressions consisting of variables, coefficients, and exponents. The degree of a polynomial is determined by the highest power of the variable within the expression. For example, in the polynomial \(f(x) = 5x^3 + 8x^2 - 4x + 3\), the highest exponent is 3, making it a third-degree polynomial.
The degree of a polynomial is significant because it tells us the maximum number of roots or zeros that the polynomial can have. In general, a polynomial of degree \(n\) can have up to \(n\) zeros, which include both real and complex solutions. This is a fundamental aspect to remember when analyzing polynomials as it sets the stage for the number of potential solutions.
The degree of a polynomial is significant because it tells us the maximum number of roots or zeros that the polynomial can have. In general, a polynomial of degree \(n\) can have up to \(n\) zeros, which include both real and complex solutions. This is a fundamental aspect to remember when analyzing polynomials as it sets the stage for the number of potential solutions.
Real Zeros
Real zeros of a polynomial are the values of \(x\) that make the polynomial equal to zero. They can be found by solving the equation \(f(x) = 0\).
To determine the possible number of positive and negative real zeros, we use Descartes' Rule of Signs. It examines the number of sign changes between consecutive non-zero coefficients in the polynomial expression:
To determine the possible number of positive and negative real zeros, we use Descartes' Rule of Signs. It examines the number of sign changes between consecutive non-zero coefficients in the polynomial expression:
- To find positive real zeros, count the sign changes in \(f(x)\).
- To find negative real zeros, substitute \(-x\) to compute \(f(-x)\), and then count the sign changes there.
Complex Zeros
Complex zeros of a polynomial include any solutions with a non-zero imaginary part. According to the Fundamental Theorem of Algebra, every polynomial equation with complex coefficients has at least one complex root (including real and imaginary zeros).
Complex roots often appear in conjugate pairs, which means if \(a + bi\) is a root, then \(a - bi\) is also a root (where \(i\) is the imaginary unit, \(i^2 = -1\)).
From the degree of the polynomial, we know the total number of real plus complex zeros must equal the polynomial degree. Considering all possible scenarios of real zeros, any shortfall in counting up to the polynomial's degree is made up by complex zeros. In our case, depending on the real zeros, the polynomial can have 0 or 2 complex zeros.
Complex roots often appear in conjugate pairs, which means if \(a + bi\) is a root, then \(a - bi\) is also a root (where \(i\) is the imaginary unit, \(i^2 = -1\)).
From the degree of the polynomial, we know the total number of real plus complex zeros must equal the polynomial degree. Considering all possible scenarios of real zeros, any shortfall in counting up to the polynomial's degree is made up by complex zeros. In our case, depending on the real zeros, the polynomial can have 0 or 2 complex zeros.
Imaginary Zeros
Imaginary zeros are a specific subset of complex zeros where the whole part (real part) is zero, and they take the form of purely imaginary numbers. These zeros arise when the real solutions do not account for the total number of zeros a polynomial should have based on its degree.
When calculating the number of imaginary zeros, we subtract the number of potential real zeros from the degree of the polynomial. In scenarios where the polynomial lacks sufficient real zeros, imaginary zeros can make up the difference.
For our polynomial \(f(x) = 5x^3 + 8x^2 - 4x + 3\), by using possible combinations of real zeros, we can deduce the number of imaginary zeros. If there are fewer real zeros than the degree indicates, the remaining zeros must be imaginary.
When calculating the number of imaginary zeros, we subtract the number of potential real zeros from the degree of the polynomial. In scenarios where the polynomial lacks sufficient real zeros, imaginary zeros can make up the difference.
For our polynomial \(f(x) = 5x^3 + 8x^2 - 4x + 3\), by using possible combinations of real zeros, we can deduce the number of imaginary zeros. If there are fewer real zeros than the degree indicates, the remaining zeros must be imaginary.
Other exercises in this chapter
Problem 3
Determine the consecutive integer values of \(x\) between which each real zero of each function is located. Then draw the graph. $$ f(x)=x^{3}-x^{2}+1 $$
View solution Problem 3
Find all of the rational zeros of each function. \(p(x)=x^{3}-5 x^{2}-22 x+56\)
View solution Problem 3
Factor completely. If the polynomial is not factorable, write prime. $$ 21-7 y+3 x-x y $$
View solution Problem 3
Find p(3) and p(-1) for each function. \(p(x)=-x^{3}+x^{2}-x\)
View solution