Problem 61 - Elementary and Intermediate Algebra

Question

Factor. If an expression is prime, so indicate. $$ 15 p^{2}-2 p q-q^{2} $$

Step-by-Step Solution

Answer

The factored form of the expression is \((3p - q)(5p + q)\).

Step 1: Identify the Coefficients and Terms

We start by identifying the coefficients and the terms in the quadratic expression: \(15p^2 - 2pq - q^2\).The terms are: \(15p^2\), \(-2pq\), \(-q^2\). The coefficients are: 15 (for \(p^2\)), -2 (for \(pq\)), and -1 (for \(q^2\)).

Step 2: Set Up for Factoring

We see that the quadratic expression is in the form of \(ax^2 + bx + c\). In our expression, \(a = 15\), \(b = -2\), and \(c = -1\). To factor this, we will use the trial and error method, searching for two binomials that multiply to obtain the original expression.

Step 3: Trial and Error Factoring

We need to find two numbers that multiply to \(a \cdot c = 15 \times (-1) = -15\) and add up to \(b = -2\). These numbers are 3 and -5 because:\(3 \times (-5) = -15\)\(3 + (-5) = -2\)

Step 4: Rewrite the Middle Term

Rewrite the middle term \(-2pq\) using the numbers found:\(15p^2 - 2pq - q^2 = 15p^2 + 3pq - 5pq - q^2\).

Step 5: Group the Terms

Group the terms to factor by grouping:\((15p^2 + 3pq) + (-5pq - q^2)\).

Step 6: Factor Each Group

Factor out the greatest common factor from each group:From \(15p^2 + 3pq\), factor out 3p:\(3p(5p + q)\)From \(-5pq - q^2\), factor out \(-q\):\(-q(5p + q)\)

Step 7: Combine the Factors

Notice both groups contain a common binomial factor \((5p + q)\). Combine:\((3p - q)(5p + q)\). This is the fully factored form of the original expression.

Key Concepts

Identifying Coefficients

To kick off the factoring process, it is essential to identify the coefficients in any quadratic expression. Consider the quadratic expression given: \(15p^2 - 2pq - q^2\). Here, the task is to pinpoint the numerical values associated with each term. These values are called coefficients.

• The term \(15p^2\) has a coefficient of 15, since it's multiplied by \(p^2\).
• The term \(-2pq\) has a coefficient of \(-2\), as it multiplies \(pq\).
• Lastly, the term \(-q^2\) has a coefficient of \(-1\), because it stands alone with \(q^2\).

Clearly identifying coefficients is the first step in recognizing the structure of the expression. Once identified, these coefficients help us understand the relationships between terms, guiding us toward the right approach to factor the expression. They lay the groundwork for subsequent steps in the factoring process.

Factoring by Grouping

Factoring by grouping is a handy method right after identifying coefficients, especially when dealing with four terms or after expanding a middle term into two. In our case, the expression is \(15p^2 + 3pq - 5pq - q^2\), which appears when we split the middle term \(-2pq\) based on numbers we identified with the trial and error method.
Breaking it down, the expression is reorganized as groups to simplify the factoring process:

• Group 1: \(15p^2 + 3pq\)
• Group 2: \(-5pq - q^2\)

This approach allows focusing on smaller sections of the expression, making it easier to factor. Each group should have a common factor that can be factored out. For Group 1, the greatest common factor is \(3p\), leading us to express it as \(3p(5p + q)\). For Group 2, the factor is \(-q\), giving us \(-q(5p + q)\).
Upon successful factoring, both groups should contain the same factor, here \((5p + q)\), which simplifies combining them into the final factored form. This method showcases how breaking down an expression makes it more approachable and less daunting.

Trial and Error Method

When other systematic methods seem cumbersome, the trial and error method can save the day. It relies on finding two numbers that meet specific multiplication and addition criteria in the expression. Here’s how it unfolds in the context of our quadratic expression, \(15p^2 - 2pq - q^2\):
We look for two numbers that, when multiplied, get us \(a \cdot c = 15 \times (-1) = -15\) and must add to \(b = -2\).

• Options might include pairs such as \(1 \times -15\), \(3 \times -5\), or \( -3 \times 5\).
• Through simple trials, the pair knowing \(3 + (-5) = -2\) satisfies the requirements efficiently.

By picking these numbers, we can replace and break down the middle term to allow for grouping. This breaks the overall problem into easier components and guides to accurate factor combinations. Though it may take a few trials, the trial and error method provides a solid pathway to the solution, ensuring all parts contribute fittingly to the end goal of factoring.