Problem 22 - College Algebra

Question

For the following exercises, factor the polynomial. $$ 2 n^{2}-n-15 $$

Step-by-Step Solution

Answer

The factored form is \((n - 3)(2n + 5)\).

Step 1: Understanding the Polynomial

The polynomial we need to factor is quadratic and given as \(2n^2 - n - 15\). We must express it in the form \((an + b)(cn + d)\) where the product of \(a\) and \(c\) is the coefficient of \(n^2\), and the product of \(b\) and \(d\) is the constant term.

Step 2: Set Up and Break Down Middle Term

Multiply the first coefficient (2) by the constant term (-15) to get -30. We need two numbers that multiply to -30 and add to -1 (the middle coefficient). These numbers are 5 and -6.

Step 3: Rewrite and Grouping

Rewrite the polynomial: \(2n^2 - n - 15 = 2n^2 + 5n - 6n - 15\). Now, group terms: \((2n^2 + 5n) + (-6n - 15)\).

Step 4: Factor by Grouping

Factor out the greatest common factor from each group. From \(2n^2 + 5n\) factor out \(n\), and from \(-6n - 15\) factor out \(-3\): \(n(2n + 5) - 3(2n + 5)\).

Step 5: Factor Out the Common Binomial Factor

Notice that \((2n + 5)\) is a common factor. Factor this out to get \((n - 3)(2n + 5)\).

Step 6: Verification

Expand \((n - 3)(2n + 5)\) to verify it equals the original polynomial: \((n - 3) \, (2n + 5) = 2n^2 + 5n - 6n - 15 = 2n^2 - n - 15\). The factorization is correct.

Key Concepts

Polynomial Factorization

Polynomial factorization involves breaking down a polynomial into a product of simpler polynomials. Think of it like unbundling a complex expression into basic building blocks. Key steps often involve identifying and applying various methods suitable for the specific type of polynomial. These methods include:

• Finding Common Factors: Look for terms that repeat across the polynomial and factor them out.
• Recognizing Patterns: Identify special factorization patterns like the difference of squares or perfect squares trinomial.

This process converts a complicated expression into its smaller components, making it easier to analyze or solve. In this exercise, we employ factorization techniques to simplify the quadratic polynomial \(2n^2 - n - 15\) into \((n - 3)(2n + 5)\). Note how we restructured and analyzed terms to express them in a product form.

Factoring by Grouping

Factoring by grouping is an essential technique particularly useful when dealing with quadratic expressions that resist simpler factoring approaches. This method involves rearranging terms and grouping them into two-factor pairs with common factors.To apply this technique, follow these steps:

• Identify terms that can be grouped together in pairs to simplify the expression.
• Factor out the greatest common factor from each group.
• If done correctly, the expression inside the brackets of each grouped term should be the same, allowing it to be factored out.

In our example, \(2n^2 - n - 15\) becomes \((2n^2 + 5n) + (-6n - 15)\), allowing us to factor by grouping, resulting in \(n(2n + 5) - 3(2n + 5)\). Finally, the common binomial \(2n + 5\) is factored out, leaving behind a simplified expression.

Quadratic Expressions

A quadratic expression is a polynomial expression where the highest degree of a variable is two, often written in the form \(ax^2 + bx + c\). These expressions are foundational in algebra, appearing frequently in a range of mathematical contexts.Key characteristics include:

• The variable is squared, forming a parabola when graphed.
• The expression can usually be factored into two linear binary expressions, helping solve for the variable.

For the quadratic expression \(2n^2 - n - 15\), identifying the coefficients helps in choosing the appropriate factoring method. Such methods enable us to reach its factored form \((n - 3)(2n + 5)\), simplifying problem-solving and analysis.