Problem 7 - College Algebra and Calculus: An Applied Approach

Question

Find all the zeros of the function and write the polynomial as a product of linear factors. $$f(x)=x^{2}+16$$

Step-by-Step Solution

Answer

The zeros of the function are \(x = 4i\) and \(x = -4i\). The polynomial as a product of linear factors is \(f(x)=(x - 4i)(x + 4i)\).

Step 1: Find the roots of the equation

First, Set \(f(x)\) equal to zero and solve for \(x\). So, solving the equation \(x^{2}+16=0\) gives us \(x^{2}=-16\). This leads to two complex solutions: \(x = 4i\) and \(x = -4i\), where \(i\) is the imaginary unit.

Step 2: Write the polynomial as a product of linear factors

The solutions to the equation are the zeros of the polynomial, and using these, the polynomial can be factored as \((x - 4i)(x + 4i)\). This is the product of linear factors for the given polynomial.

Key Concepts

Polynomial Roots

Polynomial roots are solutions to the equation formed when you set the polynomial equal to zero. In the case of the polynomial \(f(x) = x^2 + 16\), we find the roots by solving \(x^2 + 16 = 0\). To do this, we need to isolate \(x^2\), which results in \(x^2 = -16\). Since there is a negative number on the right-hand side, the solutions will involve complex numbers. What are Roots?
Roots are the values of \(x\) that make the polynomial equal to zero. They are also sometimes called zeros or solutions. For a quadratic polynomial like \(x^2 + 16\), we will typically find two roots. In this example, they are complex numbers, \(4i\) and \(-4i\).

• Solving \(x^2 + 16 = 0\), we find the roots are \(x = \pm 4i\).
• Roots can be real or complex. In this case, they are complex because of the negative square root.

Linear Factors

A polynomial can be expressed as a product of linear factors. Linear factors are simple polynomial expressions of degree one. For our polynomial \(f(x) = x^2 + 16\), we have found the roots to be \(x = 4i\) and \(x = -4i\). Factoring the Polynomial
To convert the polynomial into its linear factors, we create expressions for each root:

• For \(x = 4i\), the corresponding linear factor is \((x - 4i)\).
• For \(x = -4i\), the linear factor is \((x + 4i)\).

The polynomial \(x^2 + 16\) can thus be written as the product \((x - 4i)(x + 4i)\).
This representation is valuable for understanding the structure of polynomials and is a crucial step in many algebraic processes.

Imaginary Unit

When dealing with complex numbers, the imaginary unit is a fundamental concept. Denoted by \(i\), it is defined by the property \(i^2 = -1\). This allows us to find roots of equations that involve negative square roots, such as \(x^2 = -16\). For the polynomial \(f(x) = x^2 + 16\), using the imaginary unit enables us to solve for roots: \(x = 4i\) and \(x = -4i\). Understanding \(i\)
The introduction of \(i\) extends the real number system to the complex number system where each number has a real and an imaginary part.

• Real part: In \(4i\), the real part is 0 and the imaginary part is 4.
• Imaginary unit: The base of the imaginary part of complex numbers, crucial for calculations involving square roots of negative numbers.

Using \(i\) makes it possible to work with complex and real numbers collectively, opening up new fields of mathematics and applications in engineering and physics.