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Chapter 5: Problem 45

Multiply using the rules for the square of a binomial. $$(x+2)^{2}$$

Short Answer

Expert verified
The result of multiplying using the rules for the square of a binomial \( (x+2)^{2} \) is \( x^2 + 4x + 4 \)

Step by step solution

01

Identify the terms

Identify the terms in the binomial. Here, \( a = x \) and \( b = 2 \). Apply the formula \( (a+b)^2 = a^2 + 2ab + b^2 \).
02

Calculate a^2

\( a^2 = x^2 \)
03

Calculate 2ab

Next, calculate \( 2ab = 2*x*2 = 4x \)
04

Calculate b^2

Finally, calculate \( b^2 = (2)^2 = 4 \)
05

Sum all results

The final answer is the sum of these calculations. Therefore, \( (x+2)^2 = x^2 + 4x + 4 \)

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Key Concepts

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

Algebraic Expression
An algebraic expression is a mathematical phrase that can include numbers, variables (like \(x\) in this example), and arithmetic operations. These expressions help represent real-world problems in mathematical terms. In the exercise, \((x+2)^2\) is an algebraic expression describing the square of a binomial.
The components help in simplifying or altering the form without changing the value. Here, the expression contains the binomial \((x+2)\), which highlights the sum of \(x\) and \(2\). By understanding the structure and terms of such expressions, like constants and coefficients, we can perform various algebraic manipulations to solve or simplify problems.
Binomial Multiplication
Binomial multiplication involves multiplying two terms, typically represented in the form of \((a + b)(a + b)\). It is a basic operation in algebra used to expand expressions like the given \((x+2)^2\). The process relies on the distributive property.
When dealing with the square of a binomial, the multiplication follows a specific pattern:
  • First, you multiply the first terms of each binomial.
  • Then, you cross-multiply the inside and outside terms.
  • Finally, multiply the last terms.
Understanding this pattern aids in setting up and calculating the expanded form of an expression neatly and efficiently.
Polynomial Expansion
Polynomial expansion involves taking a simple form of a polynomial, like a squared binomial, and expanding it into a sum of terms. In this example, expanding \((x+2)^2\) transforms it into a polynomial of three terms: \(x^2 + 4x + 4\).
The process is guided by specific algebraic formulas to expand accurately:
  • Expand \(a^2\), getting \(x^2\)
  • Compute \(2ab\), resulting in \(4x\)
  • Calculate \(b^2\), which gives \(4\)
This method enables expression manipulation necessary for more complex algebraic problems. By practicing polynomial expansion, one gains fluency in handling multiple expressions in algebra.
Mathematical Formulas
The use of mathematical formulas streamlines the expansion of binomials and other expressions. For the square of a binomial, the formula \((a+b)^2 = a^2 + 2ab + b^2\) simplifies the process.
Key points about using formulas include:
  • Identifying parts of a binomial correctly (here \(a = x\), \(b = 2\)).
  • Applying the formula step-by-step to avoid mistakes.
  • Validating the final expression to ensure accuracy.
By employing formulas, we reduce the chance of errors and calculations become more efficient. Mastering the use of formulas in mathematical problems is invaluable for simplifying and solving algebraic expressions quickly.

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Most popular questions from this chapter

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. If a polynomial in \(x\) of degree 6 is divided by a monomial in \(x\) of degree \(2,\) the degree of the quotient is 4

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. $$\left(6 x^{2} y-7 x y-4\right)-\left(6 x^{2} y+7 x y-4\right)=0$$

When dividing a binomial into a polynomial with missing terms, explain the advantage of writing the missing terms with zero coefficients.

Express \(\frac{7}{8}\) as a decimal.

Solve: \(0.02(x-5)=0.03-0.03(x+7) .\) (Section 2.3 Example 5 )
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