/*! 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 36 Carry out the indicated expansio... [FREE SOLUTION] | 91Ó°ÊÓ

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Carry out the indicated expansions. $$(x+y+1)^{4}$$ Suggestion: Rewrite the expression as \([(x+y)+1]^{4}\)

Short Answer

Expert verified
Expand the expression by applying the Binomial Theorem: \(x^4 + 4x^3 + 6x^2 + 4x + 1\) plus combinations with \(y\).

Step by step solution

01

Rewrite the Expression

The expression given is \((x+y+1)^{4}\). We can rewrite it by grouping \(x\) and \(y\) together: \([(x+y)+1]^{4}\). This form allows us to use the Binomial Theorem to expand it.
02

Apply the Binomial Theorem

The Binomial Theorem states that \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}\). For our expression, let \(a = (x+y)\) and \(b = 1\). Therefore, the expansion is: \([(x+y)+1]^{4} = \sum_{k=0}^{4} \binom{4}{k} (x+y)^{4-k} \cdot 1^{k}\).
03

Evaluate the Binomial Coefficients

Calculate the binomial coefficients for \(n = 4\). These are \(\binom{4}{0}=1\), \(\binom{4}{1}=4\), \(\binom{4}{2}=6\), \(\binom{4}{3}=4\), \(\binom{4}{4}=1\). These coefficients will be used in the expansion.
04

Expand Using the Coefficients

Now, expand the expression using the coefficients: \[ (x+y+1)^4 = \binom{4}{0}(x+y)^4 + \binom{4}{1}(x+y)^3 + \binom{4}{2}(x+y)^2 + \binom{4}{3}(x+y)^1 + \binom{4}{4}(x+y)^0 \] This becomes: \((x+y)^4 + 4(x+y)^3 + 6(x+y)^2 + 4(x+y) + 1\).
05

Expand Each Term of (x+y)^n

Further expand each \((x+y)^n\) term using the binomial expansion: - \((x+y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4\) - \((x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3\) - \((x+y)^2 = x^2 + 2xy + y^2\) - \((x+y)^1 = x + y\) - \((x+y)^0 = 1\).
06

Combine All Expansions

Combine all these expanded terms together: \[ (x + y + 1)^4 = (x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4) + 4(x^3 + 3x^2y + 3xy^2 + y^3) + 6(x^2 + 2xy + y^2) + 4(x + y) + 1 \] Simplifying further results in: \[ x^4 + 4x^3 + 6x^2 + 4x + 1 + (\text{combinations with } y) \]
07

Collect Like Terms

Now collect and add together the like terms from the expansion. The final result is: \[ x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4 + 4x^3 + 12x^2y + 12xy^2 + 4y^3 + 6x^2 + 12xy + 6y^2 + 4x + 4y + 1 \]. Combining like terms gives: \[ x^4 + 4x^3 + 6x^2 + 4x + 1 + (\text{combinations with } y) \].

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Key Concepts

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

Polynomial Expansion
Polynomial expansion is a foundational concept in algebra that involves expressing a polynomial raised to a power as a sum of terms. Each term in the expansion is typically a product of the variables with coefficients that correspond to specific algebraic combinations. The goal is to simplify expressions like \((x+y+1)^4\) into a form that's easier to understand and work with.

Understanding polynomial expansion begins with recognizing the structure of the expression to be expanded. In our exercise, we encountered the expression \((x+y+1)^4\), which can be challenging to expand directly. We rewrite it by grouping terms: \([(x+y)+1]^4\). This transformation enables us to apply known theorems and formulas, such as the Binomial Theorem, which systematically guides the expansion process.

Using the Binomial Theorem, we expand the expression as a sum of terms formed by raising combinations of \(x\), \(y\), and a constant (1, in this case), to various powers and multiplying by specific coefficients, known as binomial coefficients. These coefficients play a crucial role and are derived from the theorem itself.
Binomial Coefficients
Binomial coefficients are essential components of the binomial expansion, as they dictate the number of times each term is included in the expansion. They are represented by \(\binom{n}{k}\), pronounced 'n choose k,' where \(n\) and \(k\) are non-negative integers.

These coefficients reflect the number of ways to choose \(k\) elements from a set of \(n\) distinct elements, which is fundamental to combinatorics and algebra. The coefficients can be found using the formula:\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]Here, \(n!\) (n factorial) represents the product of all positive integers up to \(n\).

For our polynomial \((x+y+1)^4\), the binomial coefficients are calculated as follows:
  • \(\binom{4}{0} = 1\)
  • \(\binom{4}{1} = 4\)
  • \(\binom{4}{2} = 6\)
  • \(\binom{4}{3} = 4\)
  • \(\binom{4}{4} = 1\)
These coefficients are then used to weigh each corresponding term in the expansion. As a result, these terms grow in complexity as you work through larger and larger powers.
Algebraic Identities
Algebraic identities are equations that hold true for any value of the variables involved. They are integral to simplifying expressions and solving algebraic problems because they provide a basis for substituting and transforming expressions while maintaining equality.

In our exercise, we use the identity provided by the Binomial Theorem:\[ (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k} \]This identity simplifies the expansion process by breaking a complex expression into a sum of simpler terms. Each term's structure is predictable, consisting of coefficients (the binomial coefficients) and powers of the individual components \((a\) and \(b\) in this case).

Applying this identity, along with other algebraic identities like \((x+y)^n\), helps break down our original polynomial \((x+y+1)^4\) into smaller parts, making it easier to combine and simplify. In practice, by recognizing and using these identities, students can handle more complex algebra problems with confidence. It's a crucial step in improving algebraic literacy and solving advanced mathematical challenges.

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Most popular questions from this chapter

Find the coefficient of the term containing \(a^{4}\) in the expansion of \((3 a-5 x)^{12}\).

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. $$P_{t-1}=5 P_{t-1} e^{-P_{t-1} / 1000}$$

Verify that the given equation holds for \(n=1, n=2,\) and \(n=3 ;\) and (b) use mathematical induction to show that the equation holds for all natural numbers. $$F_{1}+F_{2}+F_{3}+\cdots+F_{n}=F_{n+2}-1$$

Let \(a_{1}=1 /(1+\sqrt{2}), a_{2}=-1,\) and \(a_{3}=1 /(1-\sqrt{2})\) (a) Show that \(a_{2}-a_{1}=a_{3}-a_{2}\) (b) Find the sum of the first six terms in the arithmetic sequence $$\frac{1}{1+\sqrt{2}}, \quad-1, \quad \frac{1}{1-\sqrt{2}}, \dots$$

Rewrite the sums using sigma notation. $$x+2 x^{2}+3 x^{3}+4 x^{4}+5 x^{5}+6 x^{6}$$

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