Problem 85
Question
Factor each trinomial completely. $$ x^{2}+\frac{1}{2} x+\frac{1}{16} $$
Step-by-Step Solution
Verified Answer
The trinomial factors into \((x + \frac{1}{4})^2\).
1Step 1: Identify the Trinomial Structure
The given trinomial is in the form \( ax^2 + bx + c \), where \( a = 1 \), \( b = \frac{1}{2} \), and \( c = \frac{1}{16} \). This format suggests that the trinomial may be a perfect square.
2Step 2: Calculate the Square of the First Term
The square of the first term \( x^2 \) is straightforward: \( (x)^2 = x^2 \).
3Step 3: Calculate the Square of the Last Term
The last term \( \frac{1}{16} \) can be expressed as a square: \( \left( \frac{1}{4} \right)^2 = \frac{1}{16} \).
4Step 4: Check Middle Term Compatibility
If the trinomial is a perfect square, then the middle term \( b \) should be \( 2 imes \text{{(First Term)}} \times \text{{(Last Term)}} \). Calculate: \[ 2 imes x \times \frac{1}{4} = \frac{2}{4}x = \frac{1}{2}x \].The given middle term is \( \frac{1}{2}x \), which matches our calculation.
5Step 5: Express as a Perfect Square
Given the calculations confirm a perfect square, express the trinomial as the square of a binomial:\((x + \frac{1}{4})^2 = x^2 + \frac{1}{2}x + \frac{1}{16}\)
Key Concepts
Perfect Square TrinomialsAlgebraic ExpressionsFactoring Techniques
Perfect Square Trinomials
Perfect square trinomials are special types of algebraic expressions that can be expressed as the square of a binomial. When you encounter a trinomial in the form \( ax^2 + bx + c \), there is a possibility it might be a perfect square if certain conditions are satisfied. For a trinomial to be a perfect square:
- The first term \( ax^2 \) must have a perfect square, such as \( x^2 \) becoming \((x)^2\).
- The last term \( c \) should also be expressible as the square of a number, as in \( \frac{1}{16} \), which is \( \left(\frac{1}{4}\right)^2 \).
- The middle term \( b \) should equal twice the product of the base numbers of the squares of the first and last terms. For example, \( 2 \cdot x \cdot \frac{1}{4} = \frac{1}{2}x \) matches our middle term \( \frac{1}{2}x \).
Algebraic Expressions
An algebraic expression is a mathematical phrase that can include numbers, variables, and operation symbols. Understanding algebraic expressions is crucial because they are the foundation of all algebra concepts. Expressions can range from simple, like \( x + 2 \), to complex, such as polynomials.
- Variables represent unknown values that can change, making them central to forming expressions. In our exercise, \( x \) is the variable.
- Constants are fixed values, like \( \frac{1}{2} \) and \( \frac{1}{16} \) in our problem.
- Operations such as addition, subtraction, multiplication, and division connect these variables and constants to form expressions.
Factoring Techniques
Factoring is a technique used in algebra to simplify expressions or solve equations by expressing them as a product of simpler expressions. Different techniques are used depending on the type of expression you are dealing with:
- **Common Factor Extraction**: Used when all terms in the expression share a common factor. This isn't directly applicable in our example but is fundamental in other scenarios.
- **Grouping**: Grouping works well when a polynomial can be rearranged to contain common factors within subgroups.
- **Factoring Trinomials**: Identifying if a trinomial is a perfect square or employs a different recognizable pattern, like \( x^2 + bx + c = (x+p)(x+q) \), helps break it into lower degree expressions. For our exercise, recognizing it as a perfect square allows easy expression as \((x + \frac{1}{4})^2\).
- **Difference of Squares**: Applies when you have a subtractive perfect square form, though not needed for this exercise.
Other exercises in this chapter
Problem 85
Solve each equation. $$ -5 x+25=0 $$
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Fill in the chart by finding two numbers that have the given product and sum. The first column is filled in for you. $$ \begin{array}{|l|c|c|c|c|c|c|c|c|} \hlin
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Multiply. See Section 5.6. \((2 x-1)^{2}\)
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Solve each equation. $$ (2 x-3)(x+8)=(x-6)(x+4) $$
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