How do you factor completely ${x^4} + 2{x^3} - 3x - 6$?

Hint: In this question, we need to find the factors of the given polynomial. We will find the factors by grouping. Firstly, consider the first two terms and take out the common term. And then consider the remaining two terms and apply the same procedure. Then factor out the common factor. Then will get one pair as a difference of cubes and to solve this we use the identity given by,
${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})$ and simplify it to obtain the desired solution.

Complete step by step answer:
Given a polynomial of the form ${x^4} + 2{x^3} - 3x - 6$ …… (1)
We are asked to factor the polynomial given by the equation (1) completely.
Note that the highest power of the polynomial is 4, so the degree of the polynomial is equal to 4.
Now we take the first two terms as one pair and the remaining terms as another pair. We then take out the common term and then simplify it.
Now let us consider the given polynomial ${x^4} + 2{x^3} - 3x - 6$
Now we pair up it as,
$ \Rightarrow ({x^4} + 2{x^3}) - (3x + 6)$
Now we take out the common term in each pair, we get,
$ \Rightarrow {x^3}(x + 2) - 3(x + 2)$
Factoring out the common factor, we get,
$ \Rightarrow ({x^3} - 3)(x + 2)$ …… (2)
Now consider the term $({x^3} - 3)$.
We can write it as, $\left( {{x^3} - {{(\sqrt[3]{3})}^3}} \right)$
Note that this cubic term can be expressed as a difference of cubes.
We use the following identity, given by,
${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})$
Here $a = x$ and $b = \sqrt[3]{3}$.
By using the above identity, we get,
$ \Rightarrow \left( {x - \sqrt[3]{3}} \right)\left( {{x^2} + \sqrt[3]{3}x + {{\left( {\sqrt[3]{3}} \right)}^2}} \right)$
Simplifying this we get,
$ \Rightarrow \left( {x - \sqrt[3]{3}} \right)\left( {{x^2} + \sqrt[3]{3}x + \sqrt[3]{9}} \right)$
Hence substituting this in the equation (2), we get,
$ \Rightarrow \left( {x - \sqrt[3]{3}} \right)\left( {{x^2} + \sqrt[3]{3}x + \sqrt[3]{9}} \right)\left( {x + 2} \right)$

Hence by factoring the polynomial ${x^4} + 2{x^3} - 3x - 6$ we get $\left( {x - \sqrt[3]{3}} \right)\left( {{x^2} + \sqrt[3]{3}x + \sqrt[3]{9}} \right)\left( {x + 2} \right)$.

Note: Students may go wrong while factoring out the polynomials of higher degree. We need to combine the terms in the polynomial in such a way that we can take out the common term among them and simplify it.
It is important to remember the following identities.
(1) Difference of cubes : ${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})$
(2) Sum of cubes : ${a^3} + {b^3} = (a + b)({a^2} - ab - {b^2})$