/*! 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 50 Simplify the expression using th... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Simplify the expression using the binomial theorem. $$ \frac{(x+h)^{5}-x^{5}}{h} $$

Short Answer

Expert verified
The simplified expression is \(5x^4 + 10x^3 h + 10x^2 h^2 + 5x h^3 + h^4\).

Step by step solution

01

Recognize the Binomial Theorem

The binomial theorem states that \((x + h)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} h^k\). Since we need to expand \((x+h)^5\), identify that \(n=5\).
02

Expand Using Binomial Theorem

Expand \((x+h)^5\) using the binomial theorem: \( (x+h)^5 = \binom{5}{0}x^5 h^0 + \binom{5}{1}x^4 h^1 + \binom{5}{2}x^3 h^2 + \binom{5}{3}x^2 h^3 + \binom{5}{4}x^1 h^4 + \binom{5}{5}x^0 h^5 \).
03

Substitute Binomial Expansion

Substitute the expanded form into the given expression: \( \frac{(x+h)^5 - x^5}{h} = \frac{5x^4 h + 10x^3 h^2 + 10x^2 h^3 + 5x h^4 + h^5}{h} \).
04

Simplify the Expression

Factor out \( h \) from the numerator: \( \frac{h(5x^4 + 10x^3 h + 10x^2 h^2 + 5x h^3 + h^4)}{h} \).
05

Cancel Out \( h \)

Cancel \( h \) from the numerator and denominator: \( 5x^4 + 10x^3 h + 10x^2 h^2 + 5x h^3 + h^4 \).

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 91Ó°ÊÓ!

Key Concepts

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

Polynomial Expansion
In the realm of algebra, polynomial expansion is a vital concept, especially when using the binomial theorem. The binomial theorem provides a formula for expanding expressions of the form \((x + h)^n\), making it easier to handle expressions with higher powers.
To expand \((x+h)^5\), you apply the binomial theorem:
  • Each term is formed by choosing \(k\) elements from \(n\), represented as \(\binom{n}{k}\).
  • The terms are combined using increasing powers of \(h\) and decreasing powers of \(x\).
So for \((x+h)^5\), it transforms into:
  • \(\binom{5}{0}x^5 = x^5\)
  • \(\binom{5}{1}x^4h = 5x^4h\)
  • \(\binom{5}{2}x^3h^2 = 10x^3h^2\)
  • \(\binom{5}{3}x^2h^3 = 10x^2h^3\)
  • \(\binom{5}{4}xh^4 = 5xh^4\)
  • \(\binom{5}{5}h^5 = h^5\)
These terms help frame the expanded form of a binomial expression, enabling further simplification.
Simplification
Simplification of algebraic expressions involves making them more manageable by reducing their complexity. This is crucial in mathematical operations, especially when working with polynomial expansions.
Once you have the expansion \((x+h)^5 - x^5 = 5x^4h + 10x^3h^2 + 10x^2h^3 + 5xh^4 + h^5\), the goal is to simplify:
  • Cancel common factors to reduce terms to their simplest form.
  • Reduce the expression by grouping and eliminating terms.
For the given problem, after expanding, the expression is \[\frac{5x^4h + 10x^3h^2 + 10x^2h^3 + 5xh^4 + h^5}{h}\].
The simplification step is fairly strightforward:
  • Factor \(h\) from the numerator, obtaining \(h(5x^4 + 10x^3h + 10x^2h^2 + 5xh^3 + h^4)\).
  • Cancel \(h\) from both the numerator and denominator, resulting in the final simplified form \(5x^4 + 10x^3h + 10x^2h^2 + 5xh^3 + h^4\).
This makes the expression straightforward to use in further calculations.
Algebraic Expressions
Algebraic expressions represent combinations of variables and constants, organized through operations like addition, subtraction, multiplication, and division. They are foundational to algebra, allowing us to describe relationships and patterns in a compact form. In this exercise, the expression \(\frac{(x+h)^5-x^5}{h}\) is an excellent example of using algebraic expressions to simplify and solve problems through manipulation.
The manipulation of algebraic expressions often involves:
  • Understanding the structure and components of an expression.
  • Utilizing algebraic identities like the binomial theorem.
  • Applying arithmetic operations to transform and simplify the expressions.
In simplifying the given expression, we combine algebraic techniques with the expansion and reduction processes:
  • Firstly, employ the binomial theorem for expansion of \((x+h)^5\).
  • Next, carefully perform arithmetic operations to simplify the differences and fractions.
  • Finally, utilize cancellation and factorization to reduce the expression, making it more concise and clear for further use.
Mastering the manipulation of algebraic expressions is key to solving a wide range of mathematical problems.

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

A geometric design is determined by joining every pair of vertices of an octagon (see the figure). (a) How many triangles in the design have their three vertices on the octagon? (b) How many quadrilaterals in the design have their four vertices on the octagon?

Card and die experiment Each suit in a deck is made up of an ace (A), nine numbered cards \((2,3, \ldots, 10)\), and three face cards (J, Q, K). An experiment consists of drawing a single card from a deck followed by rolling a single die. (a) Describe the sample space \(S\) of the experiment, and find \(n(S)\). (b) Let \(E_{1}\) be the event consisting of the outcomes in which a numbered card is drawn and the number of dots on the die is the same as the number on the card. Find \(n\left(E_{1}\right), n\left(E_{1}^{\prime}\right)\), and \(P\left(E_{1}\right)\). (c) Let \(E_{2}\) be the event in which the card drawn is a face card, and let \(E_{3}\) be the event in which the number of dots on the die is even. Are \(E_{2}\) and \(E_{3}\) mutually exclusive? Are they independent? Find \(P\left(E_{2}\right), P\left(E_{3}\right)\), \(P\left(E_{2} \cap E_{3}\right)\), and \(P\left(E_{2} \cup E_{3}\right)\). (d) Are \(E_{1}\) and \(E_{2}\) mutually exclusive? Are they independent? Find \(P\left(E_{1} \cap E_{2}\right)\) and \(P\left(E_{1} \cup E_{2}\right)\).

Drug dosage A certain drug has a half-life of about 2 hours in the bloodstream. The drug is formulated to be administered in doses of \(D\) milligrams every 4 hours, but \(D\) is yet to be determined. (a) Show that the number of milligrams of drug in the bloodstream after the \(n\)th dose has been administered is $$ D+\frac{1}{4} D+\cdots+\left(\frac{1}{4}\right)^{n-1} D $$ and that this sum is approximately \(\frac{4}{3} D\) for large values of \(n\). (b) A level of more than 500 milligrams of the drug in the bloodstream is considered to be dangerous. Find the largest possible dose that can be given repeatedly over a long period of time.

Simplify the expression using the binomial theorem. $$ \frac{(x+h)^{4}-x^{4}}{h} $$

Rebounding ball A rubber ball is dropped from a height of 60 feet. If it rebounds approximately two-thirds the distance after each fall, use an infinite geometric series to approximate the total distance the ball travels.

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.