How do you factor \[{{x}^{4}}+2{{x}^{3}}-3x-6\]?

Hint: Take \[{{x}^{3}}\] common from the first two terms and (-3) common from the last two terms of the provided polynomial. Now, again group the common terms together. Use the formula: - \[\left( {{x}^{3}}-{{a}^{3}} \right)=\left( x-a \right)\left( {{x}^{2}}+ax+{{a}^{2}} \right)\] for further simplification. Now, check if the obtained quadratic polynomial can be factored by finding its discriminant value. If the discriminant value is positive then it can be factored and if it is negative then the quadratic polynomial cannot be factored in terms of real numbers.

Complete step by step answer:
Here, we have been provided with the polynomial \[{{x}^{4}}+2{{x}^{3}}-3x-6\] and we have been asked to factor it.
Now, in mathematics, factorization or factoring is a method of writing a number or polynomial expression as a product of several factors. For example: - let us consider a number 10, so we can write 10 as: - \[2\times 5\]. Here, 2 and 5 are called factors of 10 and the process is known as factorization. Let us take an example of a polynomial expression given as: - \[{{x}^{2}}-9\]. The considered expression can be written as: -
\[\Rightarrow {{x}^{2}}-9={{x}^{2}}-{{3}^{2}}\]
Now, using the identity, \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\], we get,
\[\Rightarrow {{x}^{2}}-9=\left( x+3 \right)\left( x-3 \right)\]
Here, \[\left( x+3 \right)\] and \[\left( x-3 \right)\] are called factors of the polynomial and the process by which these factors are found is called factorization.
Now, let us come to the question. We have the polynomial: \[{{x}^{4}}+2{{x}^{3}}-3x-6\]. Here, we will use the grouping method to get our answer. What we will do is we will group the common terms together and write the remaining terms in brackets. So, taking \[{{x}^{3}}\] common from the first two terms and (-3) common from the last two terms, we have,
\[\Rightarrow {{x}^{3}}\left( x+2 \right)-3\left( x+2 \right)\]
Now, taking \[\left( x+2 \right)\] common we get,
\[\Rightarrow \left( x+2 \right)\left( {{x}^{3}}-3 \right)\]
Here, 3 can be written as \[{{\left( 27 \right)}^{\dfrac{1}{3}}}\], so we get,
\[\begin{align}
  & \Rightarrow \left( x+2 \right)\left( {{x}^{3}}-{{\left( 27 \right)}^{\dfrac{1}{3}}} \right) \\
 & \Rightarrow \left( x+2 \right)\left[ {{x}^{3}}-{{\left( {{\left( 3 \right)}^{\dfrac{1}{3}}} \right)}^{3}} \right] \\
 & \Rightarrow \left( x+2 \right)\left[ {{x}^{3}}-{{\left( {{3}^{\dfrac{1}{3}}} \right)}^{3}} \right] \\
\end{align}\]
Applying the algebraic identity: - \[{{x}^{3}}-{{a}^{3}}=\left( x-a \right)\left( {{x}^{2}}+ax+{{a}^{2}} \right)\], we get,
\[\Rightarrow \left( x+2 \right)\left( x-{{3}^{\dfrac{1}{3}}} \right)\left( {{x}^{2}}+{{3}^{\dfrac{1}{3}}}x+{{3}^{\dfrac{2}{3}}} \right)\]
Clearly, we can see a quadratic expression in the above relation, so let us check if we can break it into more factors or not. So, we have,
\[\Rightarrow \] Discriminant = \[{{B}^{2}}-4AC\]
Here, B = coefficient of x = \[{{3}^{\dfrac{1}{3}}}\]
A = coefficient of \[{{x}^{2}}\] = 1
C = constant term = \[{{3}^{\dfrac{2}{3}}}\]
\[\Rightarrow \] Discriminant = \[{{\left( {{3}^{\dfrac{1}{3}}} \right)}^{2}}-4\times 1\times {{3}^{\dfrac{2}{3}}}=-3\times {{3}^{\dfrac{2}{3}}}\]
As we can see that the discriminant value is negative so we cannot break the quadratic expression into real factors. Therefore, the factored form can be given as: -
\[\Rightarrow {{x}^{4}}+2{{x}^{3}}-3x-6=\left( x+2 \right)\left( x-{{3}^{\dfrac{1}{3}}} \right)\left( {{x}^{2}}+{{3}^{\dfrac{1}{3}}}x+{{3}^{\dfrac{2}{3}}} \right)\]

Note: One mate note that the provided polynomial was a biquadratic expression because the highest power of the variable x is 4. Now, you may think that we would have obtained a total of four factors of this polynomial but here we got only two. Actually, the reason is that two of the factors are not real. We can find them by solving the quadratic equation but it will be in terms of \[i=\sqrt{-1}\].