Problem 23 - College Algebra with Modeling and Visualization

Question

Complete the following. (a) Find all zeros of \(f(x)\) (b) Write the complete factored form of \(f(x)\) $$ f(x)=x^{2}+25 $$

Step-by-Step Solution

Answer

Zeros are \(x = 5i\) and \(x = -5i\). Factored form is \((x - 5i)(x + 5i)\).

Step 1: Determine the Zeros of the Function

Find all values of \(x\) for which \(f(x) = 0\). We start by setting the function equal to zero:\[x^2 + 25 = 0\]Subtract 25 from both sides:\[x^2 = -25\]

Step 2: Solve for x Using Square Roots

Since \(x^2 = -25\), take the square root of both sides, remembering the properties of imaginary numbers (i.e., \(i = \sqrt{-1}\)):\[x = \pm \sqrt{-25} = \pm 5i\]This yields two solutions: \(x = 5i\) and \(x = -5i\).

Step 3: Write the Complete Factored Form

Using the zeros found, express \(f(x)\) in its factored form. We know that each zero corresponds to a factor of the form \((x - \text{zero})\):Since the zeros are \(5i\) and \(-5i\), the factors are \((x - 5i)\) and \((x + 5i)\).Thus, the factored form of the function is:\[f(x) = (x - 5i)(x + 5i)\]

Key Concepts

Factoring Polynomial Equations

Factoring polynomial equations is a method used to express the polynomial as a product of its linear factors. This is particularly useful for solving polynomial equations because it simplifies finding the roots or zeros of the function.
To factor a polynomial like the one given in the exercise, you need to first determine its zeros. Zeros are the values of the variable where the polynomial equals zero. In our example,

• The original polynomial function given is: \(f(x) = x^2 + 25\).
• The factored form of this is found by setting \(f(x) = 0\) and solving for \(x\).
• The solution reveals imaginary roots, which indicate that the factors have non-real numbers.

In the solution, we express the function as \((x - 5i)(x + 5i)\). This shows that the polynomial can be factored even though it has complex roots.

Imaginary Numbers

Imaginary numbers are numbers that, when squared, give a negative result. They extend the real number system and are particularly useful in solving equations that do not have real solutions. In our exercise,

• We encountered the expression \(x^2 = -25\).
• To solve this equation, we introduce \(i\), which is the imaginary unit defined as \(i = \sqrt{-1}\).
• The square root of \(-25\) becomes \(5i\) because \(\sqrt{-25} = \sqrt{25}\times \sqrt{-1} = 5i\).

Thus, the solutions, \(x = 5i\) and \(x = -5i\), are non-real but valid within the complex number system, illustrating the use of imaginary numbers to find zeros of functions that don't cross the real axis.

Zeros of Functions

Zeros of a function are the points where the function equals zero; these are critical in understanding the behavior of polynomial equations. Finding these zeros is a key step in factoring and solving polynomials.
For the function \(f(x) = x^2 + 25\),

• We solve \(x^2 + 25 = 0\) to find its zeros.
• The calculation yields \(x = \pm 5i\) as the zeros.

Zeros can be real or complex, and here, they are complex because the function does not intersect the x-axis. Each zero indicates a factor, enabling us to rewrite \(f(x)\) as \((x - 5i)(x + 5i)\). Understanding zeros helps in determining where a function touches or crosses within the complex plane.