/*! 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 17 Use the Binomial Theorem to expa... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 17

Use the Binomial Theorem to expand each binomial and express the result in simplified form. $$ \left(x^{2}+2 y\right)^{4} $$

Short Answer

Expert verified
The simplification of \((x^{2}+2y)^{4}\) using the Binomial Theorem is \(x^{8} + 8*x^{6}*y + 24*x^{4}*y^{2} + 32*x^{2}*y^{3} + 16*y^{4}\).

Step by step solution

01

Identify the terms and the power

The terms of our binomial are \(x^{2}\) and \(2y\) and it is raised to the power of 4. So, in our case, \(x=x^{2}\), \(y=2y\), and \(n=4\). We will be expanding \((x^{2}+2y)^4\) using the Binomial Theorem.
02

Apply the Binomial Theorem

We apply the Binomial Theorem which results in the following expression: \( (x^{2} + 2y)^{4} = \sum_{k=0}^{4} \binom{4}{k} (x^{2})^{4-k} (2y)^k\). To simplify this, we need to expand the binomial using each individual term from \(k=0\) to \(4\).
03

Expand and simplify each term

We get multiple terms, each for \(k=0\) to \(4\). Expand and simplify all the sums, then collect the like terms where possible. The sums (when expanded and simplified) will include: \(\binom{4}{0} (x^{2})^{4} (2y)^0 + \binom{4}{1} (x^{2})^{3} (2y)^1 + \binom{4}{2} (x^{2})^{2} (2y)^2 + \binom{4}{3} (x^{2})^{1} (2y)^3 + \binom{4}{4} (x^{2})^{0} (2y)^4 \), which simplifies to: \((x^{2})^{4} + 4*(x^{2})^{3}*(2y) + 6*(x^{2})^{2}*(2y)^2 + 4*(x^{2})*(2y)^3 + (2y)^4.\)
04

Final Form

Further simplifying each term we get the simplified form for our expression, which is: \(x^{8} + 8*x^{6}y + 24*x^{4}y^{2} + 32*x^{2}y^{3} + 16y^{4}\).

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 91影视!

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

Write an equation in point-slope form and slope-intercept form for the line passing through \((-2,-6)\) and perpendicular to the line whose equation is \(x-3 y+9=0 .\) (Section 2.4 Example \(2)\)

Suppose that it is a drawing in which the Powerball jackpot is promised to exceed \(\$ 700\) million. If a person purchases \(292,201,338\) tickets at \(\$ 2\) per ticket (all possible combinations), isn't this a guarantee of winning the jackpot? Because the probability in this situation is \(1,\) what's wrong with doing this?

Use the formula for the value of an annuity to solve Exercises 77鈥�84. Round answers to the nearest dollar. At age \(25,\) to save for retirement, you decide to deposit \(\$ 50\) at the end of each month in an IRA that pays \(5.5 \%\) compounded monthly. a. How much will you have from the IRA when you retire at age \(65 ?\) b. Find the interest.

Graph \(y=3 \tan \frac{x}{2}\) for \(-\pi

For the first 30 days of a flu outbreak, the number of students on your campus who become ill is increasing. Which is worse: The number of students with the flu is increasing arithmetically or is increasing geometrically? Explain your answer.
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Logic and Functions

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation
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.