# Factorize the given expression:

$4{x^3} + 4{x^2} - x - 1$

Answer

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362.7k+ views

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.

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.

Last updated date: 21st Sep 2023

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