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Chapter 1: Problem 31

Multiply the algebraic expressions using a Special Product Formula and simplify. $$(2 u+v)^{2}$$

Short Answer

Expert verified
The simplified expression is \(4u^2 + 4uv + v^2\).

Step by step solution

01

Identify the Special Product Formula

The expression \((2u + v)^2\) is a perfect square trinomial. It can be expanded using the square of a binomial formula: \((a + b)^2 = a^2 + 2ab + b^2\). Here, \(a = 2u\) and \(b = v\).
02

Calculate the Square of the First Term

Use the formula and calculate \((2u)^2 = 4u^2\). This is the first term in the expanded expression.
03

Calculate the Double Product of Both Terms

Calculate the middle term using the formula: \(2 \times 2u \times v = 4uv\).
04

Calculate the Square of the Second Term

Use the formula to find the square of the second term: \(v^2\).
05

Combine All Terms

Substitute the calculated terms into the formula: \((2u + v)^2 = 4u^2 + 4uv + v^2\). This gives the final simplified expression.

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Key Concepts

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

Perfect Square Trinomial
A perfect square trinomial is a special type of algebraic expression that arises from squaring a binomial. This expression has exactly three terms. When you see the square of a binomial, it can always be expanded into a perfect square trinomial. For example, the binomial \((a + b)^2\) expands to the trinomial \(a^2 + 2ab + b^2\).
  • The first term \(a^2\) is the square of the first part of the binomial.
  • The second term \(2ab\) accounts for twice the product of both binomial parts.
  • The third term \(b^2\) is simply the square of the second part of the binomial.
When applied to \((2u + v)^2\), it results in the trinomial \(4u^2 + 4uv + v^2\). Each of these components plays a specific role in forming the perfect square trinomial.
Binomial Expansion
Binomial expansion involves converting an expression like \((a + b)^n\) into a sum of multiple terms without parentheses. The process uses special product formulas to simplify expressions given that \(n\) is typically a small whole number. In the context of \((a+b)^2\), binomial expansion helps us systematically break down the expression into:
  • \(a^2\) for the first term squared,
  • \(2ab\) as the middle term from multiplying the terms together twice,
  • and \(b^2\) for the square of the last term.
Utilizing binomial expansion ensures each term is accounted for and provides an organized way to solve and simplify binomials like \((2u+v)^2\) into individual terms \(4u^2 + 4uv + v^2\). This simplifies understanding and computation, particularly in algebraic contexts.
Square of a Binomial
The square of a binomial is one of the most recognizable special products in algebra. It refers to multiplying a two-term expression, or binomial, by itself. The formula \((a + b)^2 = a^2 + 2ab + b^2\) is applied when squaring a binomial. This expansion involves three key components:
  • First term squared: Square the initial term \(a\) to get \(a^2\).
  • Double product of the terms: Calculate \(2ab\) by multiplying the two terms together and doubling the result.
  • Second term squared: Finally, square the second term \(b\) to get \(b^2\).
For \((2u + v)^2\), substituting \(a = 2u\) and \(b = v\) into this formula gives the expression \(4u^2 + 4uv + v^2\). The understanding of a square of a binomial simplifies working with quadratic expressions and is a powerful tool for many algebraic operations.

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Most popular questions from this chapter

Factor the expression completely. (This type of expression arises in calculus when using the "Product Rule.") $$3(2 x-1)^{2}(2)(x+3)^{1 / 2}+(2 x-1)^{3}\left(\frac{1}{2}\right)(x+3)^{-1 / 2}$$

Perform the indicated operations and simplify. $$x^{1 / 4}\left(2 x^{3 / 4}-x^{1 / 4}\right)$$

Perform the indicated operations and simplify. $$\left(x^{1 / 2}+y^{1 / 2}\right)\left(x^{1 / 2}-y^{1 / 2}\right)$$

Factor the expression completely. Begin by factoring out the lowest power of each common factor. $$x^{-3 / 2}+2 x^{-1 / 2}+x^{1 / 2}$$

Factor the expression completely. $$x^{4} y^{3}-x^{2} y^{5}$$
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