Chapter 6: Problem 2
Expand each of the following as a polynomial in powers of \(x-1\) a. \(f(x)=x^{3}-2 x^{2}+x-1\) b. \(f(x)=x^{3}+x+1\) c. \(f(x)=x^{4}\) d. \(f(x)=x^{3}-3 x^{2}+3 x\)
Short Answer
Expert verified
a. \((x-1)^3 + (x-1)^2\) b. \((x-1)^3 + 3(x-1)^2 + 4(x-1) + 3\) c. \((x-1)^4 + 4(x-1)^3 + 6(x-1)^2 + 4(x-1) + 1\) d. \((x-1)^3\)
Step by step solution
01
Identifying Binomial Expansion Need
To expand each given polynomial in terms of powers of \(x-1\), we need to transform each function so that \(x\) is re-centered around 1. This is done by substituting \(x = y+1\) where \(y=x-1\).
02
Substitute and Simplify - Part (a)
For \(f(x) = x^3 - 2x^2 + x - 1\), substitute \(x = y+1\):\((x)^3 = (y+1)^3 = y^3 + 3y^2 + 3y + 1 \-2(x)^2 = -2(y+1)^2 = -2(y^2 + 2y + 1) = -2y^2 - 4y - 2 \x = y + 1\)Gather terms:\(f(y+1) = y^3 + 3y^2 + 3y + 1 - 2y^2 - 4y - 2 + y + 1 - 1 \)Combine like terms:\(f(y+1) = y^3 + y^2 + 0(y) = (x-1)^3 + (x-1)^2\)
03
Substitute and Simplify - Part (b)
For \(f(x) = x^3 + x + 1\), substitute \(x = y+1\):\((x)^3 = (y+1)^3 = y^3 + 3y^2 + 3y + 1 \x = y + 1\) Gather terms:\(f(y+1) = y^3 + 3y^2 + 3y + 1 + y + 1 + 1 \)Combine like terms:\(f(y+1) = y^3 + 3y^2 + 4y + 3 = (x-1)^3 + 3(x-1)^2 + 4(x-1) + 3\)
04
Substitute and Simplify - Part (c)
For \(f(x) = x^4\), substitute \(x = y+1\):\((x)^4 = (y+1)^4 = y^4 + 4y^3 + 6y^2 + 4y + 1 \\)Therefore, \(f(y+1) = y^4 + 4y^3 + 6y^2 + 4y + 1 = (x-1)^4 + 4(x-1)^3 + 6(x-1)^2 + 4(x-1) + 1 \)
05
Substitute and Simplify - Part (d)
For \(f(x) = x^3 - 3x^2 + 3x\), substitute \(x = y+1\):\((x)^3 = (y+1)^3 = y^3 + 3y^2 + 3y + 1 \-3(x)^2 = -3(y+1)^2 = -3(y^2 + 2y + 1) = -3y^2 - 6y - 3 \3x = 3(y + 1)= 3y + 3\)Gather terms:\(f(y+1) = y^3 + 3y^2 + 3y + 1 - 3y^2 - 6y - 3 + 3y + 3\)Combine like terms:\(f(y+1) = y^3 = (x-1)^3\)
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Binomial Theorem
The Binomial Theorem is a powerful mathematical tool used to expand expressions raised to a power, typically of the form \((a + b)^n\). It provides a formula that makes it straightforward to determine the coefficients of the expanded terms without direct multiplication.
The theorem itself states that:\[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\]This formula involves the binomial coefficients \( \binom{n}{k} \), which are calculated using factorials.
The theorem itself states that:\[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\]This formula involves the binomial coefficients \( \binom{n}{k} \), which are calculated using factorials.
- Applications: It's extremely useful for polynomial expansions, probability, and series calculations.
- Coefficients: The coefficients are represented by the binomial coefficients, which specify how many different ways "n" can be chosen "k" times.
Substitution Method
The Substitution Method is a straightforward technique utilized in mathematical problems, particularly in polynomial expansion and re-centering processes. In the context of expanding a polynomial in powers of \((x-1)\), substitution is essential to simplify calculations and shift the polynomial's center.
Powers Substitution: In our exercise, we substitute \(x = y + 1\) where \(y = x - 1\). This transformation allows us to express the polynomial terms more conveniently as powers of \(y\). For example, replacing \(x\) with \(y+1\) makes it easier to apply the binomial expansion.
Powers Substitution: In our exercise, we substitute \(x = y + 1\) where \(y = x - 1\). This transformation allows us to express the polynomial terms more conveniently as powers of \(y\). For example, replacing \(x\) with \(y+1\) makes it easier to apply the binomial expansion.
- Purpose: This changes the polynomial into a series centered at a different point, which in this case is 1.
- Steps: Each term is modified so that all calculations happen in terms of \(y\), the transformed variable.
Polynomial Simplification
Polynomial Simplification is a process where a complex polynomial expression is simplified into a more manageable form by combining like terms and reducing the expression's overall complexity.
Key Focus:
Key Focus:
- Combine Like Terms: Terms with the same power of the variable are added or subtracted to provide a simpler expression.
- Organize Powers: Arrange terms in descending order of their degree, making the polynomial easy to read and understand.
- Reduce Coefficients: By combining like terms, coefficients often shrink, thus simplifying the overall look and computation.
Centered Polynomial
A Centered Polynomial refers to a polynomial expression that is restructured to focus around a particular point, typically making further calculations easier. In our context, the polynomials are centered at 1, demonstrated by converting them to powers of \((x-1)\).
Explanation: By substituting \(x = y+1\), where \(y = x-1\), the new polynomial expresses its terms about point 1. This is beneficial because:
Explanation: By substituting \(x = y+1\), where \(y = x-1\), the new polynomial expresses its terms about point 1. This is beneficial because:
- It simplifies derivatives: Derivatives of centered polynomials are often more straightforward.
- It provides clarity: Seeing how each term relates to a new central point offers insight, especially in series expansions.