Question

How do you use the rational root theorem to find the roots of ${{x}^{3}}+2{{x}^{2}}+3x+6=0$ ?

Answer 1

Hint : First we explain the process of grouping. We take the common divisors out to form one more common term. We then find the solution for the equation. One term always being positive we get that the other term has to be which gives the solution for the variable.

Complete step by step solution:
Factorising a polynomial by grouping is to find the pairs which on taking their common divisor out, give the same remaining number.
In case of ${{x}^{3}}+2{{x}^{2}}+3x+6$ , the grouping will be done for ${{x}^{3}}+2{{x}^{2}}$ and $3x+6$ .
We try to take the common numbers out.
For ${{x}^{3}}+2{{x}^{2}}$ , we take ${{x}^{2}}$ and get ${{x}^{2}}\left( x+2 \right)$ .
For $3x+6$ , we take 3 and get $3\left( x+2 \right)$ .
The equation becomes ${{x}^{3}}+2{{x}^{2}}+3x+6={{x}^{2}}\left( x+2 \right)+3\left( x+2 \right)$ .
Both the terms have $\left( x+2 \right)$ in common. We take that term again and get
$\begin{aligned} & {{x}^{3}}+2{{x}^{2}}+3x+6 \\\ & ={{x}^{2}}\left( x+2 \right)+3\left( x+2 \right) \\\ & =\left( x+2 \right)\left( {{x}^{2}}+3 \right) \\\ \end{aligned}$
These multiplied forms or terms can’t further be factored.
Now we can see that the value of $\left( {{x}^{2}}+3 \right)$ is always positive.
The roots of $\left( {{x}^{2}}+3 \right)$ will be imaginary. We take $\left( {{x}^{2}}+3 \right)=0$
Simplifying we get $x=\pm \sqrt{-3}=\pm i\sqrt{3}$.
The only possible choice for ${{x}^{3}}+2{{x}^{2}}+3x+6$ is $x+2=0$ .
This gives the solution of $x=-2$ .
Therefore, the factorised for form of ${{x}^{3}}+2{{x}^{2}}+3x+6=0$ is $\left( x+2 \right)\left( {{x}^{2}}+3 \right)$ .
The rational solution for the variable is $x=-2$ .
So, the correct answer is “ $x=-2$ ”.

Note : We have one more condition to check if the grouping is possible or not. If we order the individual elements of the polynomial according to their power of variables, then the multiple of end terms will be equal to the multiple of middle terms.
The given equation is in the form of ${{x}^{3}}+2{{x}^{2}}+3x+6$ . We take the multiple of two end terms and the multiple of two middle terms. Both multiplications give the result of $\left( 6 \right)\times \left( {{x}^{3}} \right)=\left( 3x \right)\times \left( 2{{x}^{2}} \right)=6{{x}^{3}}$ .