Question

How do you factor ${{x}^{4}}-{{x}^{3}}-5{{x}^{2}}-x-6$?

Answer 1

Hint: In this question we have a polynomial equation which cannot be factored further by taking out common terms from it and reducing it therefore, we will use the rational zeros theorem which uses the coefficient values of the terms and apply it in this question and find the factors of the given polynomial expression.

Complete step by step solution: We have the polynomial given as:
 $\Rightarrow {{x}^{4}}-{{x}^{3}}-5{{x}^{2}}-x-6$
Since we cannot solve this directly, we will use the rational zeros theorem to solve it.
The theorem states that if root of a polynomial is rational and is in descending order than the zero value in the form of a fraction $\pm \dfrac{p}{q}$ where $p$ is the factor of the constant and $q$ is the factor of the leading term.
In this expression, we have the coefficients as:
$1,-1,-5,-1,6$
We can see the coefficient of the leading term is $1$, the factors of this term are $\pm 1$.
The constant term is $-6$, the factors of this term are $\pm 1,\pm 2,\pm 3,\pm 6$.
Therefore, according to the theorem the rational root must be one of them:
$\Rightarrow \pm \dfrac{1}{1}=\pm 1$
$\Rightarrow \pm \dfrac{2}{1}=\pm 2$
$\Rightarrow \pm \dfrac{3}{1}=\pm 3$
$\Rightarrow \pm \dfrac{6}{1}=\pm 6$
We now have total $8$ terms and we will substitute it in the given polynomial to check for zeros.
On substituting $x=1$, we get:
$\Rightarrow 1-1-5-1-6=-12$, which is not a zero.
On substituting $x=-1$, we get:
$\Rightarrow 1+1-5+1-6=-8$, which is not a zero.
On substituting $x=2$, we get:
$\Rightarrow {{2}^{4}}-{{2}^{3}}-5\times {{2}^{2}}-2-6=20$, which is not a zero
On substituting $x=-2$, we get:
$\Rightarrow -{{2}^{4}}+{{2}^{3}}-5\times {{2}^{2}}+2-6=0$, which is zero therefore, $x=-2$ is a root.
On substituting $x=3$, we get:
$\Rightarrow {{3}^{4}}+{{3}^{3}}-5\times {{3}^{2}}-3-6=0$, which is zero therefore, $x=3$ is a root.
On substituting $x=-3$, we get:
$\Rightarrow {{3}^{4}}+{{3}^{3}}-5\times {{3}^{2}}+3-6=60$, which is not a zero.
On substituting $x=6$, we get:
$\Rightarrow {{6}^{4}}-{{6}^{3}}-5\times {{6}^{2}}-6-6=888$, which is not a zero.
On substituting $x=-6$, we get:
$\Rightarrow {{6}^{4}}+{{6}^{3}}-5\times {{6}^{2}}+6-6=1332$, which is not a zero.
From the above calculation we have found $2$ roots which are $-2$ and $3$
We can start the factorization as:
$\Rightarrow \left( x-3 \right)\left( x+2 \right)\left( a{{x}^{2}}+bx+c \right)$
On multiplying the terms, we get
$\Rightarrow \left( {{x}^{2}}-x-6 \right)\left( a{{x}^{2}}+bx+c \right)$
On expanding the brackets, we get:
$\Rightarrow a{{x}^{4}}+b{{x}^{3}}+c{{x}^{2}}-a{{x}^{3}}-b{{x}^{2}}-cx-6a{{x}^{2}}-6bx-6c$
On taking similar terms, we get:
$\Rightarrow a{{x}^{4}}+\left( b-a \right){{x}^{3}}+\left( c-6a-b \right){{x}^{2}}-\left( c+6b \right)-6c$
On comparing with the initial polynomial, we get:
$a=1,b-a=-1,c-6a-b=-5,c+6b=1,6c=6$
On simplifying, we get:
$a=1,b=0,c=1$
Therefore, the final factorization is:
$\Rightarrow \left( x-3 \right)\left( x+2 \right)\left( {{x}^{2}}+1 \right)$, which is the required answer.

Note: It is to be remembered that the solution can be also factored into the complex form as $\Rightarrow \left( x-3 \right)\left( x+2 \right)\left( x+i \right)\left( x-i \right)$ because the term $\left( {{x}^{2}}+1 \right)$ can be written as $\left( {{x}^{2}}-{{\left( \sqrt{-1} \right)}^{2}} \right)$ and expanded. it is to be remembered that a polynomial equation is a combination of variables and their coefficients, the equation given above is a $4th$ degree polynomial equation.