Problem 51 - College Algebra and Calculus: An Applied Approach

Question

Use the given zero of \(f\) to find all the zeros of \(f\). $$f(x)=x^{4}-2 x^{3}+37 x^{2}-72 x+36, \quad 6 i$$

Step-by-Step Solution

Answer

The zeros of the given function are \(x = 6i, -6i, 1\)

Step 1: Identify Conjugate

Given that \( 6i \) is a zero of \(f\), the conjugate of it, which is \( -6i \), is also a zero of \(f\). The quadratic associated with these zeros is \(x^2 + (6i - 6i)x + (6i * -6i) = x^2 + 36\).

Step 2: Divide Polynomial by Quadratic

First multiply out the function \(f\). Then divide the polynomial \(x^{4}-2 x^{3}+37 x^{2}-72 x+36\) by \(x^2 + 36\), using either polynomial long division or synthetic division. This results in the quadratic \(x^2 - 2x + 1\).

Step 3: Solve the Resulting Quadratic

We now have the quadratic \( x^2 - 2x + 1 \). Solve this quadratic for its zeros using the quadratic formula \( \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4*1*1}}{2*1} = 1 \pm 0\).

Step 4: List All Zeros

Combine the zeros from the quadratic \(x =1 \) and the conjugate pair \(x = 6i, -6i\).

Key Concepts

Complex Numbers

Complex numbers are an extension of the real numbers that include imaginary numbers. Imaginary numbers are expressed in terms of the square root of the negative one, represented as the symbol 'i'. So, a complex number can be expressed as:

• a + bi, where a and b are real numbers, and i is the imaginary unit.

The imaginary part is significant because it reveals solutions to equations that cannot be solved using only real numbers. For instance, if a polynomial equation has complex zeros, they often occur in conjugate pairs. When a polynomial has a complex zero, say, "6i," its conjugate "-6i" is also a zero. This is because complex zeros in polynomials with real coefficients appear in conjugates due to the nature of polynomial relations and symmetry in the complex plane.

Quadratic Equations

A quadratic equation is a second-degree polynomial of the form:

• \( ax^2 + bx + c = 0 \)

It is named quadratic because "quad" stands for square, which reflects the equation's highest-degree term, namely, the square of the variable. Quadratic equations can be solved using different methods:

• The Quadratic Formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
• Factoring, if applicable, when the equation reshapes into a product of binomials.
• Completing the square, which involves rearranging the terms and making a perfect square trinomial.

For the exercise at hand, after you've divided the polynomial by the quadratic equation associated with the complex roots, you're left with a simpler quadratic \( x^2 - 2x + 1 \). Applying the quadratic formula to this, you find real roots if the discriminant \( b^2 - 4ac \) is zero, indicating a perfect square, leading to a simplified root \( x = 1 \). This arithmetic step helps to precisely identify all the solutions.

Polynomial Division

Polynomial division can be compared to numerical division, but instead of numbers, you deal with variables and their exponents. It's primarily used in algebra to divide two polynomials and is handy when you need to factor polynomials or find quotient polynomials in solutions. Two common methods to perform polynomial division are:

• Long Division: Similar to traditional division, involves dividing the terms consecutively.
• Synthetic Division: A shortcut for dividing a polynomial by a linear binomial, only possible when the divisor is a first-degree polynomial of the form \( x - c \).

In the exercise, using either polynomial long division or synthetic division to divide the function \( f(x) \) by \( x^2 + 36 \) leads to the quadratic \( x^2 - 2x + 1 \). This step simplifies the process of finding all polynomial roots, as it breaks down the polynomial into manageable parts. With the correct division process, solving the polynomial becomes straightforward, and finding its solutions is efficient. This enables one to solve and understand higher-degree polynomial equations easily.