/*! 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 2 Explain in your own words how to... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 2

Explain in your own words how to square a binomial. Give an example.

Short Answer

Expert verified
To square a binomial (a + b), use the formula (a + b)^2 = a^2 + 2ab + b^2. For example, (x + 3)^2 = x^2 + 6x + 9.

Step by step solution

01

Understand a Binomial

A binomial is a polynomial with exactly two terms. For example, (a + b) is a binomial where 'a' and 'b' are any expressions.
02

State the Formula for Squaring a Binomial

The formula for squaring a binomial (a + b) is given by: (a + b)^2 = a^2 + 2ab + b^2 .
03

Apply the Formula

To demonstrate the squaring of a binomial, consider the example (x + 3). According to the formula, (x + 3)^2 = x^2 + 2(x)(3) + 3^2. This simplifies to x^2 + 6x + 9.
04

Verify the Expansion

We can verify the expansion by multiplying (x + 3) by itself: (x + 3)(x + 3). Using the distributive property, we get x(x + 3) + 3(x + 3) = x^2 + 3x + 3x + 9 = x^2 + 6x + 9.

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!

Key Concepts

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

binomial
A binomial is a type of polynomial with exactly two terms. For example, the expression \(a + b\) is a binomial, where \(a\) and \(b\) are any terms, which can be numbers, variables, or more complex expressions.
A useful feature of binomials is that they can be squared, multiplied, and divided in straightforward ways.
Understanding binomials is crucial as they are a foundation for more complex mathematical concepts.
Another typical example would be something like \(x + 2\), where \(x\) is a variable and \(2\) is a constant.
polynomial
Polynomials are mathematical expressions that consist of a sum of terms, each term including a variable raised to a non-negative integer power.
A binomial is a special kind of polynomial with exactly two terms.
For example, \(x^2 + 2x + 1\) is a polynomial with three terms - each term involving the variable \(x\) to different powers.
The degree of a polynomial is determined by the highest power of the variable present in the expression.
It's essential to identify and handle polynomials correctly, as they are frequently encountered in algebra, calculus, and many applied math problems.
distributive property
The distributive property is a key algebraic rule that helps in simplifying expressions.
It states that multiplying a sum by a number is the same as multiplying each addend by the number and then adding the products.
This can be written as \(a(b + c) = ab + ac\).
In the context of our binomial squaring, it helps to break down the multiplication of \( (x + 3)(x + 3)\).
By applying the distributive property, we get \(x(x + 3) + 3(x + 3)\), which simplifies to \(x^2 + 3x + 3x + 9 = x^2 + 6x + 9\).
Understanding and applying this property is essential for working with polynomials and binomials.
expansion
Expansion in algebra involves spreading out an expression, usually by using algebraic rules and properties.
When squaring a binomial, we use expansion to transform \( (a + b)^2 \) into its expanded form: \(a^2 + 2ab + b^2\).
This is achievable by applying the formula for squaring a binomial.
The formula can be verified by performing the multiplication using the distributive property, ensuring that our expanded (or expanded) form is accurate.
This helps in solving polynomial equations and understanding how different algebraic expressions interact.

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

Perform each division. $$ \left(2 x^{3}+x+2\right) \div(x+1) $$

Match each expression in Column I with the equivalent expression in Column II. Choices in Column II may be used once, more than once, or not at all. An Exercise \(17, x \neq 0 .\). I (a) \(-2^{-4}\) (b) \((-2)^{-4}\) (c) \(2^{-4}\) (d) \(\frac{1}{2^{-4}}\) (e) \(\frac{1}{-2^{-4}}\) (f) \(\frac{1}{(-2)^{-4}}\) II \(\mathbf{A} .8\) \(\mathbf{B} .16\) \(\mathbf{C} .-\frac{1}{16}\) \(\mathbf{D} .-8\) \(\mathbf{E} .-16\) \(\mathbf{F} . \frac{1}{16}\)

Simplify by writing each expression wth positive exponents. Assume that all variables represent nonzero real numbers. $$ \frac{\left(2 y^{-1} z^{2}\right)^{2}\left(3 y^{-2} z^{-3}\right)^{3}}{\left(y^{3} z^{2}\right)^{-1}} $$

Use scientific notation to calculate the answer to each problem. Venus is \(6.68 \times 10^{7} \mathrm{mi}\) from the sun. If light travels at a speed of \(1.86 \times 10^{5} \mathrm{mi}\) per sec, how long does it take light to travel from the sun to Venus?

Match each expression in Column I with the equivalent expression in Column II. Choices in Column II may be used once, more than once, or not at all. An Exercise \(17, x \neq 0 .\). I (a) \(x^{0}\) (b) \(-x^{0}\) (c) \(7 x^{0}\) (d) \((7 x)^{0}\) (e) \(-7 x^{0}\) (f) \((-7 x)^{0}\) II A. 0 B. 1 C. -1 D. 7 E. -7 F. \(\frac{1}{7}\)
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Statistics

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.