Problem 64 - Elementary and Intermediate Algebra

Question

Factor. See Example \(9 .\) $$v^{2}+9 v+15$$

Step-by-Step Solution

Answer

The expression \(v^2 + 9v + 15\) factors to \((v + 3)(v + 5)\).

Step 1: Identify the Coefficients

The quadratic expression given is \(v^2 + 9v + 15\), which is in the form \(ax^2 + bx + c\). Here, the coefficients are: \(a = 1\), \(b = 9\), and \(c = 15\).

Step 2: Determine the Product and Sum

We need to find two numbers that multiply to \(a \cdot c = 1 \times 15 = 15\) and add to \(b = 9\).

Step 3: Find the Pair of Numbers

Consider pairs of factors of 15: \((1, 15), (3, 5)\). The pair that adds up to 9 is \((3, 5)\).

Step 4: Rewrite the Middle Term

Express the middle term, \(9v\), using the two numbers found: \(9v = 3v + 5v\). Substitute this into the expression: \(v^2 + 3v + 5v + 15\).

Step 5: Factor by Grouping

Group the terms: \((v^2 + 3v) + (5v + 15)\). Factor each group: \(v(v + 3) + 5(v + 3)\).

Step 6: Factor Out the Common Binomial

We have a common binomial factor \((v + 3)\). Factor it out: \((v + 3)(v + 5)\). This is the fully factored form.

Key Concepts

Quadratic Expressions

In algebra, quadratic expressions are crucial for solving and understanding polynomials of degree two. A quadratic expression typically follows the format:

• \( ax^2 + bx + c \)

where:

• \( a \), \( b \), and \( c \) are constants called coefficients.
• \( x \) is the variable.
• \( a eq 0 \).

The expression represents a parabola when graphed. Quadratic expressions are foundational in mathematics as they appear in a multitude of real-world scenarios, such as calculating areas and projectile motion. Familiarity with these expressions aids in developing problem-solving skills and mathematical fluency. Quadratics can be factorable, and recognizing how to simplify or manipulate them is essential for more advanced math studies.

Factoring by Grouping

Factoring by grouping is a technique used to simplify or solve quadratic expressions. It involves breaking down a quadratic into a product of two binomials. Here's a quick rundown of how it works:1. **Identify the expression**: Start with a standard quadratic, like \( v^2 + 9v + 15 \).2. **Reorganize the middle term**: After finding two numbers that multiply to the constant term (\(c\)) and add up to the linear coefficient (\(b\)), rewrite the middle term. For example, rewrite \(9v\) as \(3v + 5v\).3. **Create groups**: Rearrange the expression into groups, \( (v^2 + 3v) + (5v + 15) \).4. **Factor each group**: Take out common factors from each set, yielding \( v(v + 3) + 5(v + 3) \). 5. **Extract the common binomial**: Finally, factor out the binomial that appears in both groups, producing \((v + 3)(v + 5)\).This method is incredibly useful when other straightforward factoring techniques (like finding roots) are not evident. It's a powerful tool for mathematicians and students alike when unraveling complex quadratic problems.

Polynomial Coefficients

Coefficients in a polynomial are the numeric factors for each term in the expression. For instance, in \( v^2 + 9v + 15 \), the coefficients are \(1\), \(9\), and \(15\). Recognizing coefficients is pivotal in managing algebraic expressions, particularly:

• **Leading Coefficient**: This is the coefficient of the term with the highest degree, and it can determine the behaviour of the graph of the polynomial.
• **Constant Term**: The stand-alone number (\(c\)), in our case, \(15\), offers insight into the value of the polynomial when all variables are zero.

Knowing how to identify and use coefficients allows us to categorize polynomials, predict their shapes, and solve them accurately. Coefficients provide ‘weight’ to each term, and they guide the method we choose for simplification, such as factoring. Understanding these basics is crucial for tackling more intricate algebraic tasks.