/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 71 The property that the product of... [FREE SOLUTION] | 91Ó°ÊÓ

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The property that the product of conjugates of the form \((a+b i)(a-b i)\) is equal to \(a^{2}+b^{2}\) can be used to factor the sum of two perfect squares over the set of complex numbers. For example, \(x^{2}+y^{2}=(x+y i)(x-y i) .\) In Exercises 71 to \(74,\) factor the binomial over the set of complex numbers. $$x^{2}+16$$

Short Answer

Expert verified
The binominal \(x^{2}+16\) can be factored as \((x + 4i)(x - 4i)\) over the set of complex numbers.

Step by step solution

01

Identify the structure

Identify that \(x^{2}+ 16\) precisely matches the pattern of \(a^{2}+b^{2}\), with \(a=x\) and \(b=4\) (as \(4^2=16\)).
02

Apply the conjugate property

Apply the property \((a+b i)(a-b i) = a^{2}+b^{2}\) to factor the binomial. Substitute \(a\) with \(x\) and \(b\) with \(4\) getting the factorised form \((x + 4i)(x - 4i)\).
03

Confirm the answer

Verify that the product of these two conjugates equals the original binomial by expanding the products: \((x + 4i)(x - 4i) = x^2 + 4xi - 4xi - 16i^2 = x^2 + 16\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Complex Conjugates
Complex conjugates form a foundational pair in the world of complex numbers. A complex number typically consists of a real part and an imaginary part, written as \( a + bi \), where \( a \) is the real part, \( b \) is the imaginary part, and \( i \) represents the square root of -1. The complex conjugate of this number is simply \( a - bi \). The significance of complex conjugates lies in their mathematical properties.

When a complex number is multiplied by its conjugate, the result is always a real number. This is because the product of a complex number and its conjugate sums to the square of the real part and adds the square of the imaginary part, effectively 'cancelling out' the imaginary parts. This leads to a crucial method for factoring expressions in the realm of complex numbers, especially when dealing with the sums of squares, which is a common case in algebra and higher mathematics.

Understanding how to work with complex conjugates is vital when factoring sums of squares because this approach will often be the most efficient solution.
Sum of Squares
The sum of squares represents a specific mathematical challenge. In real-number algebra, expressions like \(x^2 + y^2\) cannot be factored into simpler polynomials because they do not fit the pattern of a difference of squares, which would be \(x^2 - y^2\). However, this changes when we extend our perspective to complex numbers.

By introducing complex numbers into our toolkit, we're able to factor the sum of squares by using complex conjugates. The product of conjugates, as mentioned in the discussion on complex conjugates, results in the sum of the squares of the real and imaginary parts. Therefore, an expression like \(x^2 + y^2\) can be thought of as \((x + yi)(x - yi)\), which is an example of complex number factorization.

Through this lens, we can interpret the sum of squares as an extension of the difference of squares, providing a route to factor such expressions, which would otherwise seem irreducible in real-number algebra. This revelation often serves as an 'aha' moment for students grappling with complex numbers for the first time.
Complex Number Factorization
Complex number factorization is an extension of factorization techniques from real-number algebra. As we've seen with the sum of squares, the inclusion of complex numbers opens the door to new methods for breaking down algebraic expressions. This technique is particularly useful for expressions that resist factorization in the domain of real numbers.

In practice, recognizing the structure of a sum of perfect squares is the first step, as we saw in the original exercise with \(x^2 + 16\). The next step — applying the property of complex conjugates — leads to a factored form that utilizes the imaginary unit \(i\). The process boils down to identifying the real and imaginary parts, forming the conjugate pair, and then writing the expression as the product of this pair, basically the reverse of what you'd expect from expanding brackets in real-number algebra.

Keeping in mind that the imaginary unit squared \(i^2\) equals -1 aids in verifying the correctness of the factorization. This expanded toolkit not only adds dimension to a student's understanding of algebra but also provides a bridge to more advanced topics in mathematics, such as complex analysis and signal processing.

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Most popular questions from this chapter

In Exercises 51 to 60 , take square roots to solve each quadratic equation. $$(3 x+2)^{2}+100=0$$

Find a polynomial function \(P(x)\) with real coefficients that has the indicated zeros and satisfies the given conditions. Verify that \(P(x)=x^{3}-x^{2}-i x^{2}-9 x+9+9 i\) has \(1+i\) as a zero and that its conjugate \(1-i\) is not a zero. Explain why this does not contradict the Conjugate Pair Theorem.

Find the zeros of each polynomial function. If a zero is a multiple zero, state its multiplicity. $$P(x)=x^{3}+3 x^{2}-6 x-8$$

DIGITS OF PI In 1999, Professor Yasumasa Kanada of the University of Tokyo used a supercomputer to compute 206,158,430,000 digits of pi ( \(\pi\) ). (Source: Guinness World Records 2001, Bantam Books, p. \(252 .\) ) Computer scientists often try to find mathematical models that approximate the time a computer program takes to complete a calculation or mathematical procedure. Procedures for which the completion time can be closely modeled by a polynomial are called polynomial time procedures. Here is an example. A student finds that the time, in seconds, required to compute \(n \times 10,000\) digits of pi on a personal computer using the mathematical program MAPLE is closely approximated by $$\begin{aligned} T(n)=& 0.23245 n^{3}+0.53797 n^{2} \\ &+7.88932 n-8.53299 \end{aligned}$$ a. Evaluate \(T(n)\) to estimate how long, to the nearest second, the computer takes to compute 50,000 digits of \(\mathrm{pi}\). b. About how many digits of pi can the computer compute in 5 minutes? Round to the nearest thousand digits.

CANNONBALL STACKS Cannonballs can be stacked to form a pyramid with a square base. The total number of cannonballs \(T\) in one of these square pyramids is $$T=\frac{1}{6}\left(2 n^{3}+3 n^{2}+n\right)$$ where \(n\) is the number of rows (levels). If 140 cannonballs are used to form a square pyramid, how many rows are in the pyramid?

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