Problem 57 - College Algebra and Calculus: An Applied Approach

Question

Use the given zero of \(f\) to find all the zeros of \(f\). $$f(x)=8 x^{3}-14 x^{2}+18 x-9, \quad \frac{1}{2}(1-\sqrt{5} i)$$

Step-by-Step Solution

Answer

Therefore, the three roots of the polynomial \(f(x)=8 x^{3}-14 x^{2}+18 x-9\) are \(\frac{1}{2}(1-\sqrt{5}i)\), \(\frac{1}{2}(1+\sqrt{5}i)\), and \(\frac{9}{8}\).

Step 1: Identify the Complex Conjugate

Given the root \(\frac{1}{2}(1-\sqrt{5}i)\), the complex conjugate would be \(\frac{1}{2}(1+\sqrt{5}i)\). It's obtained by changing the sign of the imaginary part.

Step 2: Form Quadratic Polynomial

The quadratic polynomial is created using the original root and its complex conjugate. This is done by setting \(x-r\) for each root, then multiplying the two. \(x = \frac{1}{2}(1-\sqrt{5}i)\) and \(x = \frac{1}{2}(1+\sqrt{5}i)\) give us \((x-\frac{1}{2}(1-\sqrt{5}i))(x-\frac{1}{2}(1+\sqrt{5}i))\), which simplifies to \(x^{2} - x + 1\).

Step 3: Polynomial Division

Divide the cubic polynomial \(f(x) = 8 x^{3}-14 x^{2}+18 x-9\) by the quadratic polynomial obtained in step 2 (i.e. \(x^{2} - x + 1\)). This process provides the other factors of the polynomial. Performing the division yields the factor \(8x-9\).

Step 4: Solve for the remaining root

Equating the obtained linear factor \(8x-9\) to zero and solving for \(x\) gives the remaining root. Solving \(8x-9=0\) for \(x\) gives \(x= \frac{9}{8}\).

Key Concepts

Complex Conjugate

When working with polynomials, particularly those with complex zeros, understanding the concept of a complex conjugate is essential. For any complex number, the complex conjugate is found by simply changing the sign of the imaginary part. So, if we have a complex zero such as \( \frac{1}{2}(1-\sqrt{5}i) \), the conjugate would be \( \frac{1}{2}(1+\sqrt{5}i)\).

This is crucial in polynomial functions because complex zeros always occur in conjugate pairs when the coefficients of the polynomial are real numbers. This means that if a polynomial with real coefficients has a complex zero, its conjugate will also be a zero of the polynomial. In practice, this helps us to find the quadratic polynomial that will later divide the original polynomial, reducing the problem to finding the real zeros, which are often easier to handle.

Polynomial Division

Polynomial division, also known as long division or synthetic division when applicable, is a method of dividing a polynomial by another polynomial of lower degree. For instance, when given a cubic polynomial like \( f(x) = 8x^3 - 14x^2 + 18x - 9 \) and a quadratic polynomial such as \( x^2 - x + 1 \), you would divide the cubic by the quadratic to find the other factors.

This process is analogous to long division with numbers and helps break down complex polynomials into simpler, more manageable pieces. After the division, the result may leave us with a lower degree polynomial, which could further be factored or solved for zeros that are real and rational, aiding in our search for all zeros of the original polynomial.

Quadratic Polynomial

A quadratic polynomial is an expression of the form \( ax^2 + bx + c \) where \( a \) is not equal to zero. Quadratics are second-degree polynomials, meaning the highest power of x is 2. To find the zeros of a quadratic polynomial, we often use techniques such as factoring, completing the square, or applying the quadratic formula.

In the context of our exercise, the quadratic polynomial is formed by the given complex zero and its conjugate. Multiplying \( x - \frac{1}{2}(1-\sqrt{5}i) \) and \( x - \frac{1}{2}(1+\sqrt{5}i) \) yields a quadratic polynomial that is a factor of the original cubic polynomial, allowing for further simplification in finding all zeros. Quadratic polynomials are instrumental in solving higher-degree polynomials because once we reduce the original polynomial to a quadratic, the problem becomes much simpler.

Cubic Polynomial

A cubic polynomial is a polynomial of degree three, typically expressed as \( ax^3 + bx^2 + cx + d \) where \( a \) is nonzero. Cubic polynomials are more complex than quadratics and often require a combination of methods for finding zeros, including polynomial division, synthetic division, factoring, graphing, or numerical approximation methods.

The exercise example we have \( 8x^3 - 14x^2 + 18x - 9 \) is a cubic polynomial that we need to solve for zeros. Knowing one zero, especially a complex one, gives us a starting point to find a quadratic factor. From there, the remaining zero can be found much like solving a quadratic equation. Understanding the structure and behavior of cubic polynomials aids in predicting the number of real and complex zeros we might expect and strategizing the most efficient approach to finding all zeros.