Problem 19 - Algebra 2

Question

State the number of positive real zeros, negative real zeros, and imaginary zeros for each function. \(h(x)=4 x^{3}-6 x^{2}+8 x-5\)

Step-by-Step Solution

Answer

The function has 1 or 3 positive real zeros and 0 negative real zeros, with 2 or 0 imaginary zeros, respectively.

Step 1: Determine the Degree of the Polynomial

The function given is \(h(x)=4x^{3}-6x^{2}+8x-5\). The degree of this polynomial is 3, because the highest exponent of the variable x is 3. This indicates that the polynomial will have exactly 3 roots, considering multiplicity and both real and imaginary roots.

Step 2: Apply Descartes' Rule of Signs for Positive Real Zeros

To apply Descartes' Rule of Signs for positive real zeros, we count the number of sign changes in \(h(x)=4x^{3}-6x^{2}+8x-5\). Observing the coefficients \([+4, -6, +8, -5]\), we see three sign changes: from + to -, from - to +, and from + to -. Thus, there can be 3 or 1 positive real zeros.

Step 3: Apply Descartes' Rule of Signs for Negative Real Zeros

For negative real zeros, evaluate \(h(-x)=4(-x)^{3} - 6(-x)^{2} + 8(-x) - 5 = -4x^{3} - 6x^{2} - 8x - 5\). The sequence \([-4, -6, -8, -5]\) does not change signs, indicating there are 0 negative real zeros.

Step 4: Calculate Imaginary Zeros

The total number of roots, considering the degree, is 3. From Step 2, we know there can be 3 or 1 positive real zeros. Step 3 showed there are 0 negative real zeros. Thus, if there are 3 positive real zeros, there are 0 imaginary zeros. If there is 1 positive real zero, then there are 2 imaginary zeros.

Key Concepts

Positive Real Zeros

When dealing with polynomials, finding positive real zeros is critical. These zeros correspond to the x-values where the polynomial evaluates to zero and are positive numbers. For example, if you solve the polynomial equation and get a solution like \( x = 2 \), this means there is a positive real zero at \( x = 2 \).

To determine the number of positive real zeros, Descartes' Rule of Signs is a valuable tool. By observing the changes in sign of the coefficients of a polynomial, you can estimate the number of positive real zeros. In our exercise, considering the polynomial \( h(x) = 4x^{3} - 6x^{2} + 8x - 5 \), we notice 3 changes in sign:

• From +4 to -6
• From -6 to +8
• From +8 to -5

This means there are either 3 or 1 positive real zeros. The rule doesn't determine the exact number but gives us possible options.

Descartes' Rule of Signs

Descartes' Rule of Signs is a mathematical theorem used to estimate the number of real zeros a polynomial can have. It does so by analyzing the sign changes in the coefficients of the polynomial sequence.

Here's how it works:

• Look at the sign of each non-zero coefficient in the polynomial.
• Count changes from positive to negative or negative to positive.
• The number of sign changes gives you an upper bound on the number of positive real zeros.

For our polynomial \(4x^{3}-6x^{2}+8x-5\), the rule was applied twice:

• For the original polynomial to find the positive real zeros, resulting in 3 or 1 possible positive real zeros due to three sign changes.
• For the transformed polynomial \( h(-x) \) to look at negative real zeros, which showed no sign changes and thus no negative real zeros.

Descartes' Rule of Signs is useful because, even if it doesn't give an exact number of zeros, it narrows down the possibilities significantly.

Imaginary Zeros

Imaginary zeros occur in complex pairs when a polynomial does not factor completely into real-number solutions. These zeros are essential when the degree of the polynomial suggests more roots than real ones found.

In the polynomial \( h(x) = 4x^{3} - 6x^{2} + 8x - 5 \), we determined:

• There might be all 3 real roots (all positive), or
• There might be 1 positive real root with the remainder being imaginary zeros.

Imaginary zeros are identified when the remaining balance of expected roots from the polynomial degree cannot be accounted for by real zeros. The polynomial degree here indicates a total of 3 roots. So, if there is only 1 real solution, the remaining 2 roots must occur as complex solutions, hence 2 imaginary zeros.