Factorize the given expression:
$4{x^3} + 4{x^2} - x - 1$

Hint: Separate pair of terms together and then take the common part outside.

The given expression is $4{x^3} + 4{x^2} - x - 1$. Let its value is $y$. Then we have:
$
   \Rightarrow y = 4{x^3} + 4{x^2} - x - 1, \\
   \Rightarrow y = 4{x^2}\left( {x + 1} \right) - 1\left( {x + 1} \right), \\
   \Rightarrow y = \left( {4{x^2} - 1} \right)\left( {x + 1} \right) ...(i) \\
$
Now, we know that ${x^2} - {a^2} = \left( {x - a} \right)\left( {x + a} \right)$. Using this identity, we have $\left( {4{x^2} - 1} \right) = \left( {2x - 1} \right)\left( {2x + 1} \right)$. Putting this in equation $(i)$, we’ll get:
$
   \Rightarrow y = \left( {4{x^2} - 1} \right)\left( {x + 1} \right), \\
   \Rightarrow y = \left( {2x - 1} \right)\left( {2x + 1} \right)\left( {x + 1} \right) \\
$
Substituting the value of $y$, we have:
$ \Rightarrow 4{x^3} + 4{x^2} - x - 1 = \left( {2x - 1} \right)\left( {2x + 1} \right)\left( {x + 1} \right)$
Thus the factorized form of expression is $\left( {2x - 1} \right)\left( {2x + 1} \right)\left( {x + 1} \right)$.

Note: If we are facing any difficulty in factorizing a cubic polynomial by separating its terms, we can also find one of its roots by hit and trial. The cubic polynomial will then be divided into a linear and a quadratic expression. For example the $x = - 1$is a root of the above polynomial i.e. $4{x^3} + 4{x^2} - x - 1$. So, $\left( {x + 1} \right)$is one of its factors.
On dividing $4{x^3} + 4{x^2} - x - 1$ by $\left( {x + 1} \right)$, we will get $\left( {4{x^2} - 1} \right)$.
Thus, we have:
$ \Rightarrow 4{x^3} + 4{x^2} - x - 1 = \left( {x + 1} \right)\left( {4{x^2} - 1} \right)$
Now we can easily factorize it.