Question

Factor each polynomial completely. If the polynomial cannot be factored, say it is prime. $$ 6 x^{2}+8 x+2 $$

Short answer

The completely factored form is \(2(3x + 1)(x + 1)\).

Step-by-step solution

- Identify the polynomial

First, identify the polynomial given. The polynomial is: \( 6x^{2} + 8x + 2 \).

- Factor out the greatest common divisor (GCD)

Find the GCD of the coefficients 6, 8, and 2, which is 2. Factor out 2 from each term: \[ 6x^{2} + 8x + 2 = 2(3x^{2} + 4x + 1) \]. The polynomial inside the parenthesis is \(3x^2 + 4x + 1\).

- Factor the trinomial

To factor \(3x^{2} + 4x + 1\), we look for two numbers that multiply to \(3 \times 1 = 3\) and add to 4. These numbers are 3 and 1. Rewrite the middle term using these factors: \[ 3x^{2} + 4x + 1 = 3x^{2} + 3x + x + 1 \].

- Group and factor by grouping

Group the terms in pairs and factor out the common factor from each group: \[ 3x^{2} + 3x + x + 1 = 3x(x + 1) + 1(x + 1) \].

- Factor out the common binomial

Factor out the common binomial \((x + 1)\): \[ 3x(x + 1) + 1(x + 1) = (3x + 1)(x + 1) \]. Thus, \(3x^{2} + 4x + 1\) factors to \((3x + 1)(x + 1)\).

- Write the fully factored form

Combine the GCD and the factored form: \[ 6x^{2} + 8x + 2 = 2(3x + 1)(x + 1) \]. This is the completely factored form.

Key concepts

Factoring Trinomials

Factoring trinomials is an essential skill in algebra. A trinomial is a polynomial with three terms, such as the one from the exercise: \(6x^{2} + 8x + 2\). When factoring a trinomial, our goal is to express it as a product of simpler polynomials. This makes solving equations and simplifying expressions easier.

To factor a trinomial like \(3x^2 + 4x + 1\) from the exercise, start by finding two numbers that multiply to the product of the first and last coefficients (in this case, 3 and 1, so the product is 3) and add up to the middle coefficient (4). Here, those two numbers are 3 and 1. These numbers will help us rewrite and factor the trinomial.

Greatest Common Divisor

Before tackling the factors, first, look for the Greatest Common Divisor (GCD). The GCD is the largest number that divides all coefficients without leaving a remainder. For the polynomial \(6x^2 + 8x + 2\), the GCD is 2.

Factoring out the GCD simplifies the polynomial, making the next steps easier. In the exercise, we factored out 2 from \(6x^2 + 8x + 2\), giving us \(2(3x^2 + 4x + 1)\). Now we can focus on factoring \(3x^2 + 4x + 1\), a simpler task.

Factor by Grouping

Factoring by grouping is a technique used to factor some polynomials. After rewriting the middle term with the two numbers identified in previous steps, we group terms in pairs. For \(3x^2 + 4x + 1\), we rewrite it as \(3x^2 + 3x + x + 1\).

Group: \( (3x^2 + 3x) + (x + 1)\)
Factor common elements: \(3x(x + 1) + 1(x + 1)\). Here, \(x + 1\) is a common factor in both groups.

This lets us factor out \(x + 1\), leaving us with \((3x + 1)(x + 1)\).

Completely Factored Form

The final step is writing the polynomial in its completely factored form. For the initial polynomial \(6x^2 + 8x + 2\), we reached \(2(3x^2 + 4x + 1)\) after factoring out the GCD.

We then factored \(3x^2 + 4x + 1\) to get \((3x + 1)(x + 1)\). Combining both parts, we put the GCD and the factored terms together:
\(2(3x + 1)(x + 1)\). This is the completely factored form. This concise form is more managed and readily used in equations and further algebraic manipulations.