Problem 1
Question
Tim said that the binomial \(x^{2}+16\) can be written as \(x^{2}-16 i^{2}\) and factored over the set of complex numbers. Do you agree with Tim? Explain why or why not.
Step-by-Step Solution
Verified Answer
Yes, Tim is correct; the binomial can be factored as \((x - 4i)(x + 4i)\) over complex numbers.
1Step 1: Understand the Problem
Tim suggests rewriting the expression \(x^2 + 16\) as a difference of squares: \(x^2 - 16i^2\). This form seems to allow the expression to be factored using the difference of squares formula \(a^2 - b^2 = (a - b)(a + b)\).
2Step 2: Analyze the Rewriting
Notice that \(x^2 + 16\) can indeed be rewritten as \(x^2 - (4i)^2\) because \(16 = (4i)^2\), where \(i\) is the imaginary unit. This fits the pattern for the difference of squares.
3Step 3: Apply the Difference of Squares Formula
Using the difference of squares formula, \(a^2 - b^2 = (a - b)(a + b)\), we have \(x^2 - (4i)^2 = (x - 4i)(x + 4i)\). Thus, \(x^2 + 16 = (x - 4i)(x + 4i)\).
4Step 4: Verify the Factorization
The factorization \((x - 4i)(x + 4i)\) expands back to \(x^2 - (4i)^2 = x^2 - 16i^2\), which simplifies to \(x^2 + 16\). Thus the rewriting and factorization over the complex numbers is valid.
Key Concepts
Difference of SquaresFactorizationImaginary UnitAlgebra 2
Difference of Squares
Let's first understand what the difference of squares means. It's a particular algebraic formula used to simplify expressions where you have two perfect squares separated by a subtraction sign. The formula is:
For example, if \( x^2 - 16i^2 \) is presented, it cleverly fits into this formula which enables us to factor it into \( (x - 4i)(x + 4i) \). This shows how two apparently dissimilar expressions can reshape into products, making solving equations a lot easier in both simple and complex number contexts.
- If you have an expression in the form of \( a^2 - b^2 \), it can be factored into \((a - b)(a + b)\).
For example, if \( x^2 - 16i^2 \) is presented, it cleverly fits into this formula which enables us to factor it into \( (x - 4i)(x + 4i) \). This shows how two apparently dissimilar expressions can reshape into products, making solving equations a lot easier in both simple and complex number contexts.
Factorization
Factorization is the process of breaking down an expression into simpler "factors" that, when multiplied together, produce the original expression.
The rewriting allows us to apply formulas efficiently, hence making assessments like Tim's convenient. Complex numbers add layers, but employ standard algebraic rules with elements like \( i \) when engaging in factorization.
- It's often used to find the roots of polynomial equations or simplify expressions since smaller pieces are usually easier to work with.
The rewriting allows us to apply formulas efficiently, hence making assessments like Tim's convenient. Complex numbers add layers, but employ standard algebraic rules with elements like \( i \) when engaging in factorization.
Imaginary Unit
Now, let's dive into the concept of the imaginary unit, denoted by \( i \). The imaginary unit is special because it is defined as the square root of \(-1\).
In Algebra 2, mastery over using \( i \) allows students to tackle more sophisticated problems involving complex numbers.
- Therefore, \( i^2 = -1 \).
- Using \( i \) expands the toolset within mathematics to express quantities not feasible by real numbers alone.
In Algebra 2, mastery over using \( i \) allows students to tackle more sophisticated problems involving complex numbers.
Algebra 2
Algebra 2 typically extends concepts from Algebra 1 into more complex ideas, including a deeper understanding of polynomials and complex numbers.
This requires recognizing expressions like difference of squares even when they involve \( i \). By expanding these powers and properties encountered in Algebra 2, you navigate and solve many practical problems previously elusive.
- Students learn that expressions involving complex numbers can be handled similarly to real numbers with certain adjustments, mainly when involving the imaginary unit.
This requires recognizing expressions like difference of squares even when they involve \( i \). By expanding these powers and properties encountered in Algebra 2, you navigate and solve many practical problems previously elusive.
Other exercises in this chapter
Problem 1
a. What is the discriminant of the equation \(x^{2}+(\sqrt{5}) x-1=0 ?\) b. Find the roots of the equation \(x^{2}+(\sqrt{5}) x-1=0\) c. Do the rules for the re
View solution Problem 1
The roots of a quadratic equation with rational coefficients are \(p \pm \sqrt{q} .\) Write the equation in standard form in terms of \(p\) and \(q .\)
View solution Problem 1
Pete said that \(\sqrt{-2} \times \sqrt{-8}=\sqrt{16}=4 .\) Do you agree with Pete? Explain why or why not.
View solution Problem 1
Noah said that the solutions of Example \(1, \frac{2+\sqrt{6}}{2}\) and \(\frac{2-\sqrt{6}}{2},\) could have been written as \(\frac{2+\sqrt{6}}{2}=1+\sqrt{6}\)
View solution