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a. Is \((x-1)\) a factor of \(x^{100}-1 ?\) b. Is \((x+1)\) a factor of \(x^{100}-1 ?\) c. Is \((x-1)\) a factor of \(x^{99}-1 ?\) d. Is \((x+1)\) a factor of \(x^{99}-1 ?\) e. If \(n\) is a positive even integer, is \((x-1)\) f. If \(n\) is a positive odd integer, is \((x+1)\) a factor of \(x^{n}-1 ? \quad\) a factor of \(x^{n}-1 ?\)

Short Answer

Expert verified
a) Yes, b) Yes, c) Yes, d) No, e) Yes, f) No

Step by step solution

01

Factor Theorem

According to the Factor Theorem, \(x - a\) is a factor of a polynomial \(f(x)\) if \(f(a) = 0\).
02

Check if \(x - 1\) is a factor of \(x^{100} - 1\)

To check if \(x - 1\) is a factor of \(x^{100} - 1\), set \(x = 1\) and evaluate \(f(x) = x^{100} - 1 \). If \(f(1) = 0\), then \(x - 1\) is a factor. Calculate: \[ f(1) = 1^{100} - 1 = 1 - 1 = 0 \] Since \(f(1) = 0\), \(x - 1\) is a factor of \(x^{100} - 1\).
03

Check if \(x + 1\) is a factor of \(x^{100} - 1\)

To check if \(x + 1\) is a factor of \(x^{100} - 1\), set \(x = -1\) and evaluate \(f(x) = x^{100} - 1 \). If \(f(-1) = 0\), then \(x + 1\) is a factor. Calculate: \[ f(-1) = (-1)^{100} - 1 = 1 - 1 = 0 \] Since \(f(-1) = 0\), \(x + 1\) is a factor of \(x^{100} - 1\).
04

Check if \(x - 1\) is a factor of \(x^{99} - 1\)

To check if \(x - 1\) is a factor of \(x^{99} - 1\), set \(x = 1\) and evaluate \(f(x) = x^{99} - 1 \). If \(f(1) = 0\), then \(x - 1\) is a factor. Calculate: \[ f(1) = 1^{99} - 1 = 1 - 1 = 0 \] Since \(f(1) = 0\), \(x - 1\) is a factor of \(x^{99} - 1\).
05

Check if \(x + 1\) is a factor of \(x^{99} - 1\)

To check if \(x + 1\) is a factor of \(x^{99} - 1\), set \(x = -1\) and evaluate \(f(x) = x^{99} - 1 \). If \(f(-1) = 0\), then \(x + 1\) is a factor. Calculate: \[ f(-1) = (-1)^{99} - 1 = -1 - 1 = -2 \] Since \(f(-1) e 0\), \(x + 1\) is not a factor of \(x^{99} - 1\).
06

Check if \(x - 1\) is a factor of \(x^n - 1\) for positive even integer \(n\)

For a positive even integer \(n\), use the same reasoning as above. Set \(x = 1\) and evaluate \(f(x) = x^n - 1 \). Calculate: \[ f(1) = 1^n - 1 = 1 - 1 = 0 \] Since \(f(1) = 0\), \(x - 1\) is a factor of \(x^n - 1\) for any even \(n\).
07

Check if \(x + 1\) is a factor of \(x^n - 1\) for positive odd integer \(n\)

For a positive odd integer \(n\), use the same reasoning as above. Set \(x = -1\) and evaluate \(f(x) = x^n - 1 \). Calculate: \[ f(-1) = (-1)^n - 1 = -1 - 1 = -2 \] Since \(f(-1) e 0\), \(x + 1\) is not a factor of \(x^n - 1\) for any odd \(n\).

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Key Concepts

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

Polynomial Factorization
Polynomial factorization is the process of breaking down a polynomial into simpler products, called factors, that when multiplied together give the original polynomial. This concept is extremely useful in algebra as it simplifies complex expressions and aids in solving polynomial equations.
When we factorize polynomials, we often use the Factor Theorem to check if a certain binomial is a factor. For instance, if we have a polynomial like \(x^{100} - 1\), we might want to know if \(x - 1\) is a factor of this polynomial. According to the Factor Theorem, \(x - a\) is a factor of the polynomial \(f(x)\) if \(f(a) = 0\).
Here's a more detailed breakdown:
  • If \(f(1) = 0\), then \(x - 1\) is a factor.
  • If \(f(-1) = 0\), then \(x + 1\) is a factor.
Algebraic Expressions
An algebraic expression is a mathematical phrase that includes numbers, variables (like \(x\)), and operation symbols (such as +, -, *, /). Polynomials are a specific type of algebraic expression consisting of variables and coefficients. For example, \(x^{100} - 1\) is an algebraic expression which is also a polynomial.
Understanding algebraic expressions is essential for simplifying, factoring, and solving polynomial equations. Algebraic expressions can be simple, like \(3x + 2\), or complex, like \(4x^3 - 5x^2 + x - 12\). When working with algebraic expressions, our goal is often to simplify or to factorize them in order to solve for particular values of variables.
When we deal with exercises from textbooks, breaking down algebraic expressions step by step helps in understanding their behavior and possible factorizations.
Roots of Polynomials
The roots (or zeros) of a polynomial are the values of \(x\) for which the polynomial \(f(x)\) equals zero. For example, to check if \(x-1\) is a factor of \(x^{100} - 1\), we set \(x = 1\) and see if it makes the polynomial zero:
\[f(1) = 1^{100} - 1 = 1 - 1 = 0\]
Since \(f(1) = 0\), \(x - 1\) is a root of the polynomial. This indicates that \(x - 1\) is a factor of \(x^{100} - 1\). Similarly, by setting \(x = -1\), we can check if \(x + 1\) is a factor, since \[f(-1) = (-1)^{100} - 1 = 1 - 1 = 0.\]
Roots of polynomials give insight into the behavior and structure of the polynomial. By finding these roots, we can often factorize the polynomial into linear factors (like \(x-1\) or \(x+1\)). Knowing whether roots exist for specific values helps in solving equations and understanding the graphical representation of the polynomial function.

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Most popular questions from this chapter

Let \(a, b,\) and \(c\) represent positive real numbers, where \(a0\) c. Solve \(g(x)<0\).

A monthly phone plan costs \(\$ 59.95\) for unlimited texts and 400 min. If more than 400 min are used the plan charges an addition \(\$ 0.45\) per minute. For \(x\) minutes used, the average cost per minute \(\overline{C_{1}}(x)\) (in \$) is given by $$\overline{C_{1}}(x)=\left\\{\begin{array}{ll} \frac{59.95}{x} & \text { for } 0 \leq x \leq 400 \\ \frac{59.95+0.45(x-400)}{x} & \text { for } x>400 \end{array}\right.$$ a. Find the average cost if a customer talks for \(252 \mathrm{~min}\) \(366 \mathrm{~min},\) and \(400 \mathrm{~min} .\) Round to 2 decimal places. b. Find the average cost if a customer talks for \(436 \mathrm{~min}\) \(582 \mathrm{~min},\) and \(700 \mathrm{~min}\). Round to 2 decimal places. c. Suppose a second phone plan costs \(\$ 79.95\) per month for unlimited minutes and unlimited texts. Write an average cost function to represent the average cost \(\overline{C_{2}}(x)\) (in \$) for \(x\) minutes used. d. Find the average cost \(\overline{C_{2}}(x)\) for the second plan for \(252 \mathrm{~min}, 400 \mathrm{~min},\) and \(700 \mathrm{~min}\)

The procedure to solve a polynomial or rational inequality may be applied to all inequalities of the form \(f(x)>0, f(x)<0,\) \(f(x) \geq 0,\) and \(f(x) \leq 0 .\) That is, find the real solutions to the related equation and determine restricted values of \(x .\) Then determine the sign of \(f(x)\) on each interval defined by the boundary points. Use this process to solve the inequalities. $$ \sqrt{5-x}-7 \geq 0 $$

Explain how the solution set to the inequality \(f(x)<0\) is related to the graph of \(y=f(x)\).

A sports trainer has monthly costs of \(\$ 69.95\) for phone service and \(\$ 39.99\) for his website and advertising. In addition he pays a \(\$ 20\) fee to the gym for each session in which he trains a client. a. Write a cost function to represent the \(\operatorname{cost} C(x)\) for \(x\) training sessions. b. Write a function representing the average \(\operatorname{cost} \bar{C}(x)\) for \(x\) sessions. c. Evaluate \(\bar{C}(5), \bar{C}(30),\) and \(\bar{C}(120)\). d. The trainer can realistically have 120 sessions per month. However, if the number of sessions were unlimited, what value would the average cost approach? What does this mean in the context of the problem?

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