Problem 8 - 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. $$f(x)=x^{2}+36$$

Step-by-Step Solution

Answer

The zeros of the function \(f(x) = x^{2} + 36\) are \(x = \pm 6i\). The function can be written as a product of linear factors as \(f(x) = (x - 6i)(x + 6i)\).

Step 1: Setting the Function Equal to Zero

The first thing to do is to set the function equal to zero. This is done because we are looking for the zeros of the function, meaning the values of \(x\) that make the function equals zero. So, we start with the equation \(x^{2} + 36 = 0\).

Step 2: Solving for x

Subtract 36 from both sides to get \(x^{2} = -36\). Now you have \(x^{2} = -36\). This is a simple equation to solve for \(x\), but the square root of negative number will result in imaginary numbers.

Step 3: Dealing with Imaginary Numbers

Taking the square root of both sides of the equation gives \(x = \pm \sqrt{-36}\). We can turn the negative under the square root into an imaginary number by taking the square root of -1, which is denoted by \(i\), and the square root of 36, which is 6. Therefore, \(x = \pm 6i\). These are the zeros of the function.

Step 4: Writing the Polynomial as a Product of Linear Factors

The polynomial can be written as a product of linear factors using the zeros found in the previous step. In this case, the linear factors are \(x - 6i\) and \(x + 6i\). So, \(f(x) = (x - 6i)(x + 6i)\). Checking it with the original polynomial, we find that when we multiply these linear factors we indeed get back the original polynomial.

Key Concepts

Zeros of a function

The zeros of a function are the values of \( x \) that make the function equal to zero. In mathematical terms, if \( f(x) \) is a function, its zeros are the solutions to the equation \( f(x) = 0 \). Finding zeros is crucial because they tell us where the function crosses or touches the x-axis on a graph.

• To find zeros, set the function equal to zero. For \( f(x) = x^2 + 36 \), set it as \( x^2 + 36 = 0 \).
• Solving this gives \( x^2 = -36 \), which is key in understanding that not all solutions are real numbers.

In this example, since we end up with a negative number under the square root, we encounter imaginary numbers. This transition highlights how zeros can sometimes lead us into the realm of complex numbers.

Polynomial factorization

Polynomial factorization involves expressing a polynomial as a product of simpler polynomials (or factors). For quadratic polynomials like \( x^2 + 36 \), we often try to factor them into linear factors.

• After finding the zeros \( x = \pm 6i \), these become the roots for our linear factors.
• Each zero \( a \) leads to a factor \( (x - a) \).

Thus, \( f(x) = x^2 + 36 = (x - 6i)(x + 6i) \). When you multiply these linear factors, you should end up with the original polynomial. This factorization allows for a deeper understanding of the polynomial's roots and their nature, whether they are real or imaginary.

Imaginary Numbers

Imaginary numbers arise when dealing with the square roots of negative numbers. In standard form, imaginary numbers are expressed as multiples of \( i \), where \( i \) is the imaginary unit defined by \( i^2 = -1 \).

• In the equation \( x^2 = -36 \), the square root operation gives \( x = \pm 6i \).
• This demonstrates that when the square of a number is negative, the solutions involve imaginary numbers.

Imaginary numbers are a critical component of complex numbers, which are numbers of the form \( a + bi \) where \( a \) and \( b \) are real numbers. They allow mathematicians to solve equations that have no real solutions and expand the power of algebra in many branches of science and engineering.