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Use the Binomial Theorem to factor the expression. $$16 z^{4}+32 z^{3}+24 z^{2}+8 z+1$$

Short Answer

Expert verified
Answer: \((2z + 1)^4\)

Step by step solution

01

Identify the common factor in every term

Observe the given expression: $$16z^4 + 32z^3 + 24z^2 + 8z + 1$$ Notice that every term's coefficient is a power of 2. Also, each term has a power of \(z\) that decreases with each term; the last term's power of \(z\) being 0.
02

Find a binomial expression

Based on the observation in step 1, we can rewrite the expression as: $$ \left( 2^4 z^4 \right) + \left( 2 \cdot 2^4 z^3 \right) + \left( {4\choose2} 2^4 z^2 \right) + \left( {4\choose3} 2^4 z \right) + \left( {4\choose4} 2^4 \right)$$ Now, we can identify the binomial expression. Notice that each term matches the general form of a binomial expansion term: \({n\choose k} a^{n-k} b^k\), where \(n = 4\), \(a = 2z\), and \(b = 1\).
03

Apply the Binomial Theorem

The Binomial Theorem states that: $$(a+b)^{n} = \sum_{k=0}^{n} {n\choose k} a^{n-k}b^k$$ Now, we know from step 2 that \(a = 2z\), \(b = 1\), and \(n = 4\). We plug these values into the Binomial Theorem formula: $$(2z + 1)^4 = \sum_{k=0}^{4} {4\choose k} (2z)^{4-k}1^k$$ Therefore, the factored expression is: $$16z^4 + 32z^3 + 24z^2 + 8z + 1 = (2z + 1)^4$$

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Key Concepts

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

Factoring Polynomials
When dealing with algebraic expressions, factoring polynomials involves breaking them down into simpler, non-divisible polynomials that multiply together to give the original polynomial. It's like finding the building blocks of an expression. For instance, the polynomial \(16z^4 + 32z^3 + 24z^2 + 8z + 1\) might appear daunting at first, but by finding common factors and patterns, we can simplify it. In our exercise, we noticed that coefficients were powers of 2, hinting at a binomial expression lying underneath.

To efficiently factor such expressions, looking out for the greatest common factor, difference of squares, sum or difference of cubes, or patterns resembling the binomial expansion can be crucial. The process often involves reverse engineering the expansion of expressions, especially when spotting a pattern related to the binomial theorem, as seen in our initial expression.
Binomial Expansion
The binomial expansion refers to expressing a binomial, \((a + b)^n\), as a sum of terms involving the base numbers 'a' and 'b' raised to varying exponents, multiplied by specific coefficients. These coefficients are known as the binomial coefficients. For example, \((a+b)^2 = a^2 + 2ab + b^2\).

The process becomes increasingly complicated as 'n' increases, prompting the use of the Binomial Theorem, which provides a systematic method to expand any binomial raised to a positive integer exponent. It's like having a recipe for creating a delicious meal, where the ingredients are mixed in precisely the right quantities. The theorem states that
\[(a+b)^n = \sum_{k=0}^{n} {n\choose k} a^{n-k}b^k\],
which breaks down a seemingly complex problem into a series of simpler steps, allowing us to expand binomials quickly and accurately, without manually multiplying the binomial 'n' times.
Combinatorial Coefficients
Combinatorial coefficients, commonly presented as \({n\choose k}\), represent the number of ways to choose 'k' items from 'n' options without regard to the order of selection. In mathematics, these are called the binomial coefficients because they appear in the binomial expansion as multipliers for each term. They can also be referred to as 'combinations' in probability and statistics.

These coefficients are not only crucial in algebra but also have vast applications in combinatorics, probability theory, and other areas of mathematics. Calculating these values can be done using factorials, where \({n\choose k} = \frac{n!}{k!(n-k)!}\), with '!' denoting the factorial operation. Understanding these coefficients helps in appreciating the underlying uniformity and pattern within the binomial expansion, making it easier to work with and understand expressions within the context of the Binomial Theorem.
Exponents and Powers
Exponents and powers are the shorthand in mathematics to express repeated multiplication of a number by itself. An exponent denotes how many times a base number is multiplied, formatted as \(a^n\), where 'a' is the base and 'n' is the exponent. It succinctly describes large values without writing out long multiplication strings and is essential in many areas of math, including polynomial expression and binomial expansion.

Understanding how to manipulate exponents with laws of exponents is fundamental. For example, when multiplying terms with the same base, you add the exponents, and when raising a power to a power, you multiply the exponents. Recognizing the patterns in powers of numbers, as seen with the coefficients in our polynomial factoring example, can also accelerate the process of factoring by revealing hidden binomial structures. This foundational knowledge assists significantly when expanding binomials, as it helps in organizing terms and simplifying complex expressions.

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Most popular questions from this chapter

In Exercises \(13-22,\) one term and the common ratio r of a geometric sequence are given. Find the sixth term and a formula for the nth term. $$a_{1}=\pi, r=\frac{1}{5}$$

Deal with prime numbers. A positive integer greater than 1 is prime if its only positive integer factors are itself and 1. For example, 7 is prime because its only factors are 7 and \(1,\) but 15 is not prime because it has factors other than 15 and 1 (namely, 3 and 5 ). Find the first five terms of the sequence. \(a_{n}\) is the \(n\) th prime integer larger than \(10 .\left[\text {Hint}: a_{1}=11 .\right]\)

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Prove that for any positive integer \(n\) $$\begin{array}{l}\left(\begin{array}{l}n \\\0\end{array}\right)-\left(\begin{array}{l}n \\\1\end{array}\right)+\left(\begin{array}{l}n \\ 2\end{array}\right)-\left(\begin{array}{l}n \\\3\end{array}\right)+\left(\begin{array}{l}n \\\4\end{array}\right)-\cdots \\\\+(-1)^{k}\left(\begin{array}{l}n \\\k\end{array}\right)+\cdots+(-1)^{n}\left(\begin{array}{l}n \\\n\end{array}\right)=0\end{array}$$

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