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