What happens when the quadratic expression becomes more complicated? Suppose instead of the form \(ax^2 + bx + c \), the quadratic takes on the form \(a(f(x))^2 + b(f(x)) + c\), where there is a repetitive expression found within the trinomial. There are a couple of methods that can be used to factor these types of trinomials. One is to expand and simplify the expression, followed by factoring. The other method is to use substitution to factor, and then simplify the factors. Follow this second method in the next example.

 Example  3


Factor the expression \(3\left( {2x + 5} \right)^2  + 3\left( {2x + 5} \right) - 6\).

Step 1: Substitute a new variable to represent the expression \(2x + 5\).

Let \(u = 2x + 5\)
\(3\left( {2x + 5} \right)^2 + 3\left( {2x + 5} \right) - 6 = 3u^2 + 3u - 6\)

Step 2: Factor the simplified version of the quadratic expression.

\(3u^2 + 3u - 6 = 3\left( {u^2 + u - 2} \right) = 3\left( {u + 2} \right)\left( {u - 1} \right)\)

Step 3: Replace the substituted variable with the original expression.

\(3\left( {u + 2} \right)\left( {u - 1} \right) = 3\left[ {\left( {2x + 5} \right) + 2} \right]\left[ {\left( {2x + 5} \right) - 1} \right]\)

Step 4: Simplify the factors.

\(\begin{align}
 3\left[ {\left( {2x + 5} \right) + 2} \right]\left[ {\left( {2x + 5} \right) - 1} \right] &= 3\left( {2x + 7} \right)\left( {2x + 4} \right) \\
  &= 3\left( {2x + 7} \right)\left( 2 \right)\left( {x + 2} \right) \\
  &= 6\left( {2x + 7} \right)\left( {x + 2} \right) \\
 \end{align}\)

Notice that after simplifying the factors, a new GCF was found.