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Carry out the indicated expansions. $$(x+\sqrt{2})^{8}$$

Short Answer

Expert verified
The expanded form of \((x+\sqrt{2})^8\) is: \[x^8 + 8\sqrt{2}x^7 + 56x^6 + 112\sqrt{2}x^5 + 280x^4 + 224\sqrt{2}x^3 + 224x^2 + 64\sqrt{2}x + 16\]

Step by step solution

01

Understand the Expansion Method

The problem requires expanding \((x+\sqrt{2})^8\). The most straightforward method is using the binomial theorem, which expresses any power of a binomial \((a+b)^n\) as a series of terms: \[(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\] where \(\binom{n}{k}\) is the binomial coefficient.
02

Identify the Components

For the expression \((x+\sqrt{2})^8\), identify \(a = x\), \(b = \sqrt{2}\), and \(n = 8\). These components will be plugged into the binomial theorem formula.
03

Write the Expansion Formula

Using the binomial theorem, expand \((x+\sqrt{2})^8\) as: \[(x+\sqrt{2})^8 = \sum_{k=0}^{8} \binom{8}{k} x^{8-k} (\sqrt{2})^k\] This step sets up the equation we will use to find each term in the expansion.
04

Compute the Binomial Coefficients

Calculate each binomial coefficient \(\binom{8}{k}\) for \(k = 0, 1, 2, ..., 8\). The binomial coefficients are: \(\binom{8}{0} = 1\), \(\binom{8}{1} = 8\), \(\binom{8}{2} = 28\), \(\binom{8}{3} = 56\), \(\binom{8}{4} = 70\), \(\binom{8}{5} = 56\), \(\binom{8}{6} = 28\), \(\binom{8}{7} = 8\), \(\binom{8}{8} = 1\).
05

Determine Each Term of the Expansion

Substitute \(k\) from 0 to 8 into the formula: \(\binom{8}{k} x^{8-k} (\sqrt{2})^k\) and calculate each term: - \(k=0\): \(1 \cdot x^8\)- \(k=1\): \(8 \cdot x^7 \cdot \sqrt{2}\)- \(k=2\): \(28 \cdot x^6 \cdot 2\)- \(k=3\): \(56 \cdot x^5 \cdot 2\sqrt{2}\)- \(k=4\): \(70 \cdot x^4 \cdot 4\)- \(k=5\): \(56 \cdot x^3 \cdot 4\sqrt{2}\)- \(k=6\): \(28 \cdot x^2 \cdot 8\)- \(k=7\): \(8 \cdot x \cdot 8\sqrt{2}\)- \(k=8\): \(1 \cdot 16\)
06

Simplify Each Term

Now, simplify each term: - \(1 \cdot x^8 = x^8\)- \(8 \cdot x^7 \cdot \sqrt{2} = 8\sqrt{2} x^7\)- \(28 \cdot x^6 \cdot 2 = 56x^6\)- \(56 \cdot x^5 \cdot 2\sqrt{2} = 112\sqrt{2} x^5\)- \(70 \cdot x^4 \cdot 4 = 280x^4\)- \(56 \cdot x^3 \cdot 4\sqrt{2} = 224\sqrt{2} x^3\)- \(28 \cdot x^2 \cdot 8 = 224x^2\)- \(8 \cdot x \cdot 8\sqrt{2} = 64\sqrt{2} x\)- \(1 \cdot 16 = 16\)
07

Write the Final Expanded Expression

Combine all the simplified terms into the final expanded expression: \[x^8 + 8\sqrt{2}x^7 + 56x^6 + 112\sqrt{2}x^5 + 280x^4 + 224\sqrt{2}x^3 + 224x^2 + 64\sqrt{2}x + 16\] This is the expansion of \((x+\sqrt{2})^8\).

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Key Concepts

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

Binomial Coefficients
When expanding expressions such as \((x + \sqrt{2})^8\), binomial coefficients play a vital role. These coefficients are part of the Binomial Theorem, a fundamental concept in algebra. They determine the weight of each term in the polynomial expansion.
The binomial coefficient, denoted by \(\binom{n}{k}\), is a way to choose \(k\) elements from a set of \(n\) elements without regard to the order of selection. The formula for calculating a binomial coefficient is:
  • \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\)
Here, \(n!\) (n factorial) is the product of all positive integers up to \(n\). These coefficients essentially determine how many times each term appears in the expansion.
In the example provided, for \((x + \sqrt{2})^8\), the calculated binomial coefficients for each term from \(k = 0\) to \(k = 8\) are: 1, 8, 28, 56, 70, 56, 28, 8, 1. These numbers are crucial for determining the individual terms in our expanded expression.
Polynomial Expansion
Polynomial expansion involves expressing a power of a binomial as a sum of terms. Each term is a product of a binomial coefficient, a power of the first term, and a power of the second term.
Using the Binomial Theorem, the polynomial expansion of \((a + b)^n\) can be expressed as:
  • \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\)
For instance, expanding \((x + \sqrt{2})^8\) involves identifying \(a = x\) and \(b = \sqrt{2}\), and substituting into the formula:
  • \((x+\sqrt{2})^8 = \sum_{k=0}^{8} \binom{8}{k} x^{8-k} (\sqrt{2})^k\)
Each term is computed by varying \(k\) from 0 to 8, evaluating the powers of \(x\) and \(\sqrt{2}\), and applying the corresponding binomial coefficient. This results in a polynomial with multiple terms, each term representing a different component of the expansion.
Algebraic Manipulation
Algebraic manipulation is the process of simplifying and rearranging expressions. During polynomial expansion using the binomial theorem, we need to manipulate the terms to simplify the final expression.
In the example \((x + \sqrt{2})^8\), each term calculated includes a coefficient, a power of \(x\), and a power of \(\sqrt{2}\). For instance, one term is calculated as \(8 \times x^7 \times \sqrt{2} = 8\sqrt{2} x^7\).
Simplification requires careful multiplication and understanding of exponent rules, especially when dealing with square roots or higher powers. The expression must be consistently rearranged from terms involving the highest power of \(x\) down to the constant term.

  • The term \(56 \times x^5 \times 2\sqrt{2}\) simplifies to \(112\sqrt{2} x^5\).
  • Each term's simplification might involve a numerical factor and a factor involving \(\sqrt{2}\).
Ultimately, all terms are combined to form the fully expanded and simplified polynomial expression. This manipulation ensures clarity and correctness in expressing the polynomial.

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Most popular questions from this chapter

An important model that is used in population biology and ecology is the Ricker model. The Canadian biologist William E. Ricker introduced this model in his paper Stock and Recruitment (Journal of the Fisheries Research Board of Canada, \(11(1954) 559-623\) ). For information on Ricker himself, see the web page The general form of the Ricker model that we will use here is defined by a recursive sequence of the form \(P_{0}=\) initial population at time \(t=0\) \(P_{t}=r P_{t-1} e^{-k P_{t-1}} \quad\) for \(t \geq 1,\) and where \(r\) and \(k\) are positive constants (a) Suppose that the initial size of a population is \(P_{0}=300\) and that the size of the population at the end of year \(t\) is given by$$P_{t}=5 P_{t-1} e^{-P_{t-1} / 1000} \quad(t \geq 1)$$ Use a graphing utility to compute the population sizes through the end of year \(t=5 .\) (As in Example 5, round the final answers to the nearest integers.) Then use the graphing utility to draw the population scatter plot for \(t=0,1, \ldots, 5 .\) Describe in complete sentences how the size of the population changes over this period. Does the population seem to be approaching an equilibrium level? (b) Using a graphing utility, compute the sizes of the population in part (a) through the end of the year \(t=20\) and draw the corresponding scatter plot. Note that the population seems to be approaching an equilibrium level of \(1609(\text { or } 1610)\) (c) Determine the equilibrium population algebraically by solving the following equation for \(P_{t-1} .\) For the final answer, use a calculator and round to the nearest integer. $$P_{t-1}=5 P_{t-1} e^{-P_{t-1} / 1000}$$

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