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When faced with an algebraic expression, the first stride in factoring often involves finding the Greatest Common Factor (GCF). The GCF is the largest factor that divides all the terms in the expression. In the context of our exercise, the trinomial given was \(16s^3 - 28s^2 + 6s\). To find the GCF for this trinomial:

• List the factors of each coefficient and include the variables' exponents: 16, 28, and 6 share a GCF of 2, and considering the variable \(s\), which appears in each term, \(s\) is also part of the GCF.
• Thus, the GCF is the product of these common factors, which is \(2s\).
• Next, divide each term by \(2s\) to factor it out from the trinomial.

Doing so simplifies the expression, reducing its complexity and preparing it for further factoring steps.

Quadratic expressions are polynomials of the second degree, generally in the form \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable. After factoring out the GCF in our trinomial, we obtained \(8s^2 - 14s + 3\), which is a quadratic expression. The next phase involves determining if this quadratic can be factored further.

To factor a quadratic expression, we aim to write it as the product of two binomials. Ideally, we are looking for two numbers that multiply to give the product of \(a\) and \(c\) (here, \(8 \times 3 = 24\)) and add up to \(b\) (here, -14). If such numbers exist, the quadratic can be written as \( (as + m)(cs + n) \) where \( m \times n = ac \) and \( m + n = b \) in our general form.

Algebraic factorization is a method to simplify expressions and solve equations by expressing a polynomial as the product of its factors. The principles of algebraic factorization guide us through splitting the trinomial into binomial factors that will multiply back to the original quadratic expression.

For the quadratic expression \(8s^2 - 14s + 3\), we look for a pair of numbers that fulfills the multiplication and addition requirements we set in the previous section on quadratic expressions. The process, which can be trial-and-error, led us to the binomials \(4s - 1\) and \(2s - 3\) for our problem. Multiplying these binomials will return the original expression, thus confirming correct factorization.

The final factored form of the original trinomial, \(16s^3 - 28s^2 + 6s\), is then \(2s(4s - 1)(2s - 3)\), showcasing how the concepts of GCF, quadratic expressions, and algebraic factorization work in tandem to break down increasingly complex algebra problems.