Factor the expression \(3\left( {x^2 + 7x + 10} \right)^2 + 4\left( {x^2 + 7x + 10} \right) - 7\)

Let \(u = x^2 + 7x + 10\)
\(3\left( {x^2 + 7x + 10} \right)^2  + 4\left( {x^2 + 7x + 10} \right) - 7 = 3u^2 + 4u - 7\)

Factor the new expression.

\(3u^2 + 4u - 7 = \left( {3u + 7} \right)\left( {u - 1} \right)\)

Substitute \(u = x^2 + 7x + 10\), and simplify.

\(\begin{align}
 \left( {3u + 7} \right)\left( {u - 1} \right) &= \left[ {3\left( {x^2 + 7x + 10} \right) + 7} \right]\left[ {\left( {x^2 + 7x + 10} \right) - 1} \right] \\
  &= \left( {3x^2 + 21x + 30 + 7} \right)\left( {x^2 + 7x + 9} \right) \\
  &= \left( {3x^2 + 21x + 37} \right)\left( {x^2 + 7x + 9} \right) \end{align}\)


Note that these trinomials cannot be further factored.