Chapter 3: Problem 55
Factor the polynomial function \(f(x)=x^{2}-i\)
Short Answer
Expert verified
The polynomial \( f(x) = x^2 - i \) factors as \((x - (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i))(x + (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i))\).
Step by step solution
01
Identify the Polynomial and Imaginary Component
The given polynomial function is \( f(x) = x^2 - i \), where \( i \) is the imaginary unit, \( i = \sqrt{-1} \). Our task is to factor this polynomial into a product of two linear complex expressions.
02
Recognize the Structure for Factoring
The polynomial \( x^2 - i \) can be compared to the structure \( a^2 - b^2 = (a+b)(a-b) \), which allows us to use its difference of squares to factor. Here, we need to express \( i \) as a square.
03
Express \( i \) as a Square
We need to write \( i \) in a squared form. Notice that \( i = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i)^2 \). This gives us a way to apply the difference of squares.
04
Factor the Polynomial
Now using the difference of squares form \( x^2 - (i) = x^2 - (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i^2) \), we can factor \( f(x) = x^2 - i \) as \((x - (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i))(x + (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i))\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Complex Numbers
Complex numbers are a fascinating area of mathematics, opening a world beyond the real number line. A complex number is the combination of a real part and an imaginary part. The general form of a complex number is \( a + bi \), where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit. Complex numbers allow us to extend our math toolbox for solving equations that do not have real solutions, like the square root of a negative number.
Key points about complex numbers include:
Key points about complex numbers include:
- They can be plotted on the complex plane, similar to how we plot coordinates on a graph.
- The real part represents the horizontal axis and the imaginary part represents the vertical axis.
Imaginary Unit
The imaginary unit, denoted as \( i \), is a mathematical concept used to extend the real numbers. It is defined by the property \( i^2 = -1 \). This definition allows us to deal with roots of negative numbers, which are not possible within the scope of real numbers.
In our example exercise, \( i \) was used in the given polynomial function \( f(x) = x^2 - i \), representing a simple yet powerful tool in algebra. Here is what makes \( i \) important:
In our example exercise, \( i \) was used in the given polynomial function \( f(x) = x^2 - i \), representing a simple yet powerful tool in algebra. Here is what makes \( i \) important:
- It helps solve equations that have no real solution, enriching the set of possible solutions.
- Combining \( i \) with real numbers leads to complex numbers, expanding the real number line into a complex plane.
- Operations involving \( i \) follow classic algebraic rules, albeit adjusted for the defining condition of \( i^2 = -1 \).
Difference of Squares
The difference of squares is a technique used frequently in algebra to simplify expressions and solve equations. It applies to expressions of the form \( a^2 - b^2 \), which can be factored into \( (a + b)(a - b) \). This is a specific case of the more general formula known as "difference of two squares."
In our polynomial \( x^2 - i \), recognizing the structure reminiscent of a difference of squares enables simplification even when dealing with complex numbers:
In our polynomial \( x^2 - i \), recognizing the structure reminiscent of a difference of squares enables simplification even when dealing with complex numbers:
- The first step involves expressing the constant component as a square term, as we did with \( i \).
- Then, factor the polynomial using the difference of squares formula \( a^2 - b^2 \).
- This leads to linear factors that reveal the polynomial's roots or solutions, even in the complex plane.