Problem 36 - Algebra A Combined Function

Question

Factor each trinomial completely. See Examples 1 through 5 . \(3 n^{2}+20 n+5\)

Step-by-Step Solution

Answer

The trinomial \(3n^2 + 20n + 5\) factors as \((n + 5)(3n + 1)\).

Step 1: Identify the Trinomial Structure

The given trinomial is in the form \(ax^2 + bx + c\), where \(a = 3\), \(b = 20\), and \(c = 5\). Our goal is to factor this trinomial into a product of two binomials.

Step 2: Check for a Greatest Common Factor (GCF)

First, check if there is a greatest common factor other than 1 among the coefficients \(3\), \(20\), and \(5\). Since there is no common factor greater than 1, there is no GCF to factor out.

Step 3: Use the AC Method

Multiply \(a\) and \(c\) together: \(3 \times 5 = 15\). We need two numbers that multiply to 15 and add up to \(b = 20\).

Step 4: Find Suitable Numbers

List the pairs of factors of 15: \(1 \times 15\) and \(3 \times 5\). Among these, the pair that adds up to 20 is \(1\) and \(15\).

Step 5: Rewrite the Middle Term

Rewrite the middle term \(20n\) using the numbers found: \(3n^2 + 1n + 15n + 5\). This allows us to split the trinomial into four terms.

Step 6: Factor by Grouping

Group the terms: \((3n^2 + 1n) + (15n + 5)\). Factor out the common factor from each group: \(n(3n + 1) + 5(3n + 1)\).

Step 7: Factor Out the Common Binomial Factor

Notice the common binomial \((3n + 1)\), factor this out: \((n + 5)(3n + 1)\).

Step 8: Verify the Factorization

Expand \((n + 5)(3n + 1)\) to check: \(3n^2 + n + 15n + 5 = 3n^2 + 20n + 5\). The factored expression matches the original trinomial, ensuring that our factorization is correct.

Key Concepts

Greatest Common Factor (GCF)

In any factoring problem, the first step is always to check for a Greatest Common Factor (GCF) among the coefficients of the terms. The GCF is the largest factor that divides all the coefficients without leaving a remainder.
This is an essential step because factoring out the GCF simplifies the polynomial, making it easier to handle. For the trinomial shown in the exercise, the coefficients are 3, 20, and 5. We check for a GCF other than 1. Since none of these numbers share a common factor greater than 1, there is no GCF to factor out.
This means we proceed to the next step in factoring, but always remember to check for the GCF first as it can greatly simplify more complex polynomials.

AC Method

The AC Method is a systematic approach to factor trinomials, particularly when the coefficient of the squared term (abla aabla) is not equal to 1, which is the case with our trinomial.
Here's how it works:

• Multiply the coefficient of the squared term (\(a = 3\)) by the constant term (\(c = 5\)). Here, you get: \(3 \times 5 = 15\).
• Look for two numbers that multiply to give this result (15) and add up to the linear coefficient (\(b = 20\)).

Examining the factors of 15, we find the pair (1, 15) meets these conditions. Using these numbers, we split the middle term into two terms.
This helps convert the trinomial into a form that can be factored using grouping, thus simplifying the process of deriving the binomial factors.

Binomial Factoring

Once we have the trinomial rewritten as a four-term polynomial, binomial factoring often follows through a process called grouping. The exercise has already demonstrated this method using the AC Method.
Here's how it proceeds:

• Group the terms: Separate the expression into two groups, resulting in: \((3n^2 + 1n) + (15n + 5)\).
• Find the common factor in each group: \(n\) is common in the first group and 5 in the second, giving: \(n(3n + 1) + 5(3n + 1)\).
• Factor out the common binomial: Each group includes the binomial factor \((3n + 1)\), enabling us to factor it out, resulting in: \((n + 5)(3n + 1)\).

This factorization shows the original trinomial expressed as a product of two binomials. This step connects all the previous processes, ensuring the equation is simplified and correctly factored.
Verifying the factorization by expanding back to the initial terms ensures accuracy.