28x^{3} + 15x^{2} + 2x

First notice that there is greatest common factor (gcf) among all three term in the equation, that being x. So we first factor out an x, then we are left to factor a simple quadratic equation:

28x^{3} + 15x^{2} + 2x = x(28x^{2} + 15x + 2)

= x( )( )

We know that the terms on the left hand side of each set of parentheses have to multiply out to give 28x^{2} and the terms on the right hand side of each set of parentheses have to multiple out to be 2. Additionally, the sum of the products of the two outer terms and the two inner terms must be 15x. So we need to find a set of factor for 28 and 2 that will satisfy these conditions...

Factors of 28: 1·28, 2·14, 4·7

Factors of 2: 1·2

Since the only set of factors for 2 are 1 and 2, we arrive at the following:

x(28x^{2} + 15x + 2) = x( + 1)( + 2)

It is obvious that if we try to use 1 and 28 to factor 28, the sum of the products will be much greater than 15. Likewise, 2 and 14 will also yield a sum larger than 15. Thus, 4 and 7 is our best choice of factors for 28:

x(28x^{2} + 15x + 2) = x(4x + 1)(7x + 2)

Be careful as to which set of parentheses you place these terms. For example, if we interchanged the 4x and 7x we would get a different answer:

x(7x + 1)(4x + 2) = x(28x^{2} + 14x + 4x + 2)

= x(28x^{2} + 18x + 2) ≠ x(28x^{2} + 15x + 2)

Thus, the trinomial in question is factored out to:

28x^{3} + 15x^{2} + 2x = x(4x + 1)(7x + 2)