Problem 22 - College Algebra with Corequisite Support

Question

For the following exercises, factor the polynomial. \(2 n^{2}-n-15\)

Step-by-Step Solution

Answer

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

Step 1: Identify the Polynomial Structure

The polynomial is a quadratic expression of the form \(ax^2 + bx + c\). Here, \(a = 2\), \(b = -1\), and \(c = -15\). Our goal is to factor this expression into the product of two binomials.

Step 2: Set Up the Factorization

We are looking for two numbers that multiply to \(a \times c\) which is \(2 \times (-15) = -30\), and add up to \(b = -1\).

Step 3: Find the Correct Pair

The pair of numbers that multiply to \(-30\) and sum to \(-1\) are \(5\) and \(-6\). This is because \(5 \times (-6) = -30\) and \(5 + (-6) = -1\).

Step 4: Rewrite the Middle Term

Rewrite the middle term \(-n\) using the pair found in Step 3. Thus, the polynomial becomes \(2n^2 + 5n - 6n - 15\).

Step 5: Group the Terms

Group the terms into two pairs: \((2n^2 + 5n)\) and \((-6n - 15)\).

Step 6: Factor Each Group

Factor out the common factors in each group. For \(2n^2 + 5n\), factor out \(n\), resulting in \(n(2n + 5)\). For \(-6n - 15\), factor out \(-3\), resulting in \(-3(2n + 5)\).

Step 7: Factor the Entire Expression

Notice both groups contain the common binomial \((2n + 5)\). Thus, factor this out to get \((2n + 5)(n - 3)\).

Step 8: Verify the Solution

Expand \((2n + 5)(n - 3)\) to check the factorization is correct. Multiplying, we get \(2n^2 + 5n - 6n - 15 = 2n^2 - n - 15\), which matches the original expression.

Key Concepts

Quadratic Expression

A quadratic expression is a type of polynomial that holds terms of the second degree. In simpler terms, a quadratic expression typically takes the format \(ax^2 + bx + c\), where:

• \(a\) is the coefficient of the squared term \(x^2\),
• \(b\) is the coefficient of the linear term \(x\),
• and \(c\) is the constant term.

The defining feature of a quadratic expression is its highest degree, which is two, making the graph of these equations typically appear as a parabola.
Quadratic expressions are common in various fields such as physics, engineering, and finance because they can model wide-ranging phenomena. For example, the trajectory of a thrown ball can be represented by a quadratic equation.
Understanding how to work with them is crucial for solving many mathematical problems, which often involve finding factors or roots of these expressions.

Polynomial Factoring

Polynomial factoring is a mathematical process aimed at breaking down polynomials into simpler "factors." These factors, when multiplied together, yield the original polynomial. Factoring is akin to breaking numbers into prime constituents but applied to expressions instead of numerical values.
The essence of factoring is to express a polynomial as a product of other polynomials, usually of lower degrees. This is especially useful as it offers insights into the polynomial's roots, solutions, and behavior on a graph.In the quadratic expression \(2n^2-n-15\), the goal was to simplify this polynomial into the product of two binomials. Multiplying these binomials back should equal the original expression, showcasing how factoring gives a pathway to simplification and problem-solving in algebra. Learning and mastering polynomial factoring is beneficial in advanced algebra, calculus, and many practical applications.How do factors relate to expressions? Simply put, knowing the factors can help solve equations related to the expression, find intercepts on graphs, and understand polynomial properties like roots and their multiplicity.

Factoring by Grouping

Factoring by grouping is a technique used when a polynomial does not abide by straightforward factoring methods or simple recognition of common factors. This method involves arranging terms into groups that can be easily factored.
Using the previous step-by-step solution, we break down the expression \(2n^2-n-15\) by the following steps:

• Rewrite the expression as four terms (as shown: \(2n^2 + 5n - 6n - 15\)). This is achieved by finding two numbers that multiply to \(a \times c\) and sum to \(b\).
• Next, group terms into pairs. Here, \((2n^2+5n)\) and \((-6n-15)\) are paired together.
• Factor out the greatest common factor (GCF) from each pair, resulting in \(n(2n+5)\) from the first and \(-3(2n+5)\) from the second.
• Finally, notice the common binomial \(2n+5\) and factor this out, giving the final factored form, \((2n+5)(n-3)\).

This technique is powerful as it reveals the polynomial’s factors by intuitively rearranging and simplifying terms, thus offering a strategy that can handle more complex expressions than typical factoring.