/*! 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 11 Factor \(b^{x}\) out of the foll... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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

Short Answer

Expert verified
The factored expression of \(b^{2 x}+b^{x+1}\) is \(b^{x}(b^{x}+b)\)

Step by step solution

01

Identify the common factor

Look at each term of the equation \(b^{2 x}+b^{x+1}\) and identify the common factor. Here, \(b^{x}\) is the common factor for both terms.
02

Factor out the common factor

Factor out \(b^{x}\) from the equation. The result will be \(b^{x}(b^{x}+b)\). This is obtained by dividing each of the original terms by the common factor \(b^{x}\).
03

Check the answer by multiplying out

Verify the answer by multiplying what you factored back into the parentheses. If you distribute \(b^{x}\) back through the parentheses, it should equal your original expression, \(b^{2 x}+b^{x+1}\). After verifying, we can see that the factored form is correct.

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.

Common Factor
Understanding the concept of a common factor is crucial when simplifying mathematical expressions. In algebra, factoring expressions involves breaking down equations into simpler components, and identifying a common factor plays an integral role in this process.

A common factor is a term that can be evenly divided from all other terms in an expression without any remainder. This term, which can be a number, variable, or a combination of both, is shared among the components of the expression. For instance, if we look at the expression from the exercise \(b^{2x}+b^{x+1}\), we can observe that \(b^x\) appears in both terms and hence, can be considered as a common factor.

Once you locate the common factor, you can use it to factor the expression by essentially 'taking it out' of each term. This can be helpful to simplify the expression and make it easier to work with in solving equations or evaluating algebraic formulas. For example: \(b^x\) as a common factor in the given expression leads to a simplified form \(b^x(b^x+b)\).
Exponential Expressions
Exponential expressions are a recurrent theme in algebra that involve variables with exponents. They are written as \(b^n\), where \(b\) is the base and \(n\) is the exponent, indicating that \(b\) is multiplied by itself \(n\) times.

Understanding the laws of exponents can be incredibly helpful for manipulating these expressions. Some basic laws include the product rule \(b^{m} \cdot b^{n} = b^{m+n}\), the quotient rule \(\frac{b^{m}}{b^{n}} = b^{m-n}\), and the power rule \(\left(b^{m}\right)^{n} = b^{mn}\).

In the exercise \(b^{2x}+b^{x+1}\), we see two different exponential expressions. To factor out the common factor \(b^x\), we must recognize how exponents behave. For example, \(b^{x+1}\) is the same as \(b^x \cdot b\), and factoring out \(b^x\) leaves us with just \(b\) as a reminder of this term. Similarly, factoring out \(b^x\) from \(b^{2x}\) leaves \(b^x\) as it is equivalent to \(b^x \cdot b^x\).
Distributive Property
The distributive property is a foundational principle in algebra that allows us to multiply a single term by each term inside a bracket (or parentheses) separately. It is expressed as \(a(b+c) = ab + ac\). This property is particularly important when factoring expressions or multiplying polynomials.

When an expression like \(b^{2x}+b^{x+1}\) is given, you can apply the distributive property in reverse to factor out the common factor, \(b^x\), by dividing each term of the expression by \(b^x\) to see what would be inside the parentheses. The distributive property is also used to verify if the factoring is correct by multiplying the factored form back out and comparing it to the original expression.

For example, in our exercise, when you multiply \(b^x\) by \(b^x + b\), each term inside the parenthesis should be multiplied by \(b^x\), which gives us back our original expression \(b^{2x} + b^{x+1}\), affirming that our factored form is accurate and confirming the power of the distributive property in simplifying and verifying algebraic expressions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In the middle of the \(1994-95\) academic year, in the middle of the week, in the middle of the day, there was a bank robbery and subsequent shootout in the middle of Harvard Square. Throughout the afternoon the news spread by word-of- mouth. Suppose that at the time of the occurrence 30 people know the story. Every 15 minutes each person who knows the news passes it along to one other person. Let \(N(t)\) be the number of people who know at time \(t\). (a) Make a table with time in one column and \(N(t)\) in the other. Identify the pattern and write \(N\) as a function of \(t\). (b) If you've written your equation for \(N(t)\) with \(t\) in minutes, convert to hours. If you've done it in hours, convert to minutes. Make a table to check your answers. (It's easy to make a mistake the first time you do this.)

Use the tangent line approximation of \(e^{x}\) at \(x=0\) to approximate \(e^{-1} .\) Is your answer larger than \(e^{-1}\) or smaller?

Exploratory: Which grows faster, \(2^{x}\) or \(x^{2}\) ? (a) Using what you know about these two functions and experimenting numerically and graphically, guess the following limits: i. \(\lim _{x \rightarrow \infty} x 2^{-x}\) ii. \(\lim _{x \rightarrow-\infty} \frac{x^{2}}{2^{x}} \quad\) iii. \(\lim _{x \rightarrow \infty} \frac{x^{2}}{2^{x}} \quad\) iv. \(\lim _{x \rightarrow \infty} \frac{2^{x}}{x^{2}}\) (b) For \(|x|\) large, which function is dominant, \(2^{x}\) or \(x^{2} ?\) Would you have answered differently if we looked at \(3^{x}\) and \(x^{3}\) instead?

If the statement is always true, write "True;" if the statement is not always true, produce a counterexample. In these problems, \(a\) and \(b\) are positive constants. $$ \sqrt{\frac{a^{2 x}}{b^{-2 x}}}=(a b)^{x} $$

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.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.