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Chapter 9: Problem 12

Factor \(b^{x}\) out of the following expressions. Check your answer by multiplying out. $$ 3 b^{2 x+1}-4 b^{2 x-1} $$

Short Answer

Expert verified
The factorized form of the expression is \(b^{x}(3b^{x+1}-4b^{x-1})\).

Step by step solution

01

Identify Common Factor

The algebraic expression given here is \(3 b^{2x+1}-4 b^{2x-1}\). The common factor in both terms is \(b^{x}\). Note that \(b^{2x+1}\) is equivalent to \(b^{x} \times b^{2x}\), and \(b^{2x-1}\) is equivalent to \(b^{x} \times b^{x-1}\).
02

Factor out the Common Term

Factor out \(b^{x}\). Replace each term of the original expression by what鈥檚 left when you take out the common factor. This gives the factored form as \(b^{x}(3b^{x+1}-4b^{x-1})\).
03

Expand the factored expression

To verify the factoring, go through the process in reverse by multiplying \(b^{x}\) to each term within the parentheses. This gives \(3b^{2x+1}-4b^{2x-1}\). Hence, the correctness of the solution is validated.

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Key Concepts

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

Understanding Algebraic Expressions
Algebraic expressions are mathematical phrases representing numbers and operations. They include variables, constants, and operators. In the context of the expression given, we noticed terms like \(3b^{2x+1}\) and \(-4b^{2x-1}\). Here, \(b\) is a variable raised to different powers, and coefficients such as 3 and -4 are multipliers of these terms. Understanding how to manipulate these expressions is crucial in algebra. It involves simplifying them or factoring them to reveal their underlying structure. Recognizing patterns and common terms will make it easier to work with these expressions.
Identifying the Common Factor
A common factor is a term that appears in multiple parts of an expression. It allows you to simplify by pulling out the shared term from each part. The goal is to rewrite the expression in a more manageable form. To find the common factor, look for terms that all the parts share. In this exercise, the expression \(3 b^{2x+1} - 4 b^{2x-1}\) both have \(b^{x}\) as a common factor. Once identified, the common factor can be factored out. This simplifies the expression and reveals a new perspective on its form.
Multiplying Out the Expression
Once an expression is factored, multiplying out is a reverse process to verify correctness. It involves distributing the extracted common factor back into the expression. For our example, to check if \(b^{x}(3b^{x+1} - 4b^{x-1})\) is correct, multiply \(b^{x}\) with each term inside the parentheses:
  • \(b^{x} \times 3b^{x+1} = 3b^{2x+1}\)
  • \(b^{x} \times -4b^{x-1} = -4b^{2x-1}\)
By performing these multiplications, we return to the original expression \(3 b^{2x+1} - 4 b^{2x-1}\), thus validating the factorization.
Verifying the Factoring Process
Verifying factoring involves ensuring that the factorization process has been done correctly. After factoring, the expression should be expandable back to its original form. This checks your work and ensures accuracy. In our case, once we factor out \(b^{x}\), multiplying it back into the resulting terms should reproduce the original \(3 b^{2x+1} - 4 b^{2x-1}\). Always take this step in factoring to confirm the results. It not only confirms that the solution is accurate but also reinforces understanding of the process's mechanics.

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Most popular questions from this chapter

In the Exploratory Problems you approximated the derivatives of \(2^{x}, 3^{x}\), and \(10^{x}\) for various values of \(x\), and, after looking at your results, you conjectured about the patterns. Now, using the definition of the derivative of \(f\) at \(x=a\), we return to this, focusing on the function \(f(x)=5^{x}\). (a) Using the definition of the derivative of \(f\) at \(x=a\), $$ f^{\prime}(a)=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h} $$ give an expression for \(f^{\prime}(0)\), the slope of the tangent line to the graph of at \(x=0\). (b) Show that for the function \(f(x)=5^{x}\), the difference quotient, \(\frac{f(x+h)-f(x)}{h}\), is equal to \(f(x) \cdot \frac{f(h)-f(0)}{h}\). (c) Using the definition of derivative, $$ f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} $$ conclude that the derivative of \(f(x)=5^{x}\) is $$ f^{\prime}(0) \cdot f(x) $$ Notice that you have now proven that the derivative of \(5^{x}\) is proportional to \(5^{x}\), with the proportionality constant being the slope of the tangent line to \(5^{x}\) at \(x=0\). $$ f^{\prime}(x)=f^{\prime}(0) \cdot f(x) $$ (d) Approximate the slope of the tangent line to \(5^{x}\) at \(x=0\) numerically.

Factor \(b^{x}\) out of each of the following expressions. (a) \(3 b^{x}-b^{2 x}\) (b) \((3 b)^{x}-b^{x+2}\) (c) \(b^{3 x / 2}-b^{2 x-1}\)

For Problems 1 through 9, simplify the following expressions. $$ \frac{\left(a^{-x+1} b\right)^{3}}{\left(a^{2} b^{3}\right)^{x}} $$

Find the equation of the line tangent to \(f(x)=e^{x}\) at \(x=1\).

Suppose that in a certain scratch-ticket lottery game, the probability of winning with the purchase of one card is 1 in 500 , or \(0.2 \%\); hence, the probability of losing is \(100 \%-0.2 \%=99.8 \%\). But what if you buy more than one ticket? One way to calculate the probability that you will win at least once if you buy \(n\) tickets is to subtract from \(100 \%\) the probability that you will lose on all \(n\) cards. This is an easy calculation; the probability that you will lose two times in a row is \((99.8 \%)(99.8 \%)=\) (a) What is the probability that you will win at least once if you play three times? (b) Find a formula for \(P(n)\), the percentage chance of winning at least once if you play the game \(n\) times. (c) How many tickets must you buy in order to have a \(25 \%\) chance of winning? A \(50 \%\) chance? (d) Does doubling the number of tickets you buy also double your chances of winning? (e) Sketch a graph of \(P(n) .\) Use \([0,100]\) as the range of the graph. Explain the practical significance of any asymptotes.
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