Question

How do you find all the real and complex roots of ${{x}^{3}}+{{x}^{2}}+x+2=0$ ?
(a) Factorization
(b) Guessing the roots
(c) Changing the variable
(d) None of the above

Answer 1

Hint: To find all the real and complex roots, we are to try to factorize the equation given as, ${{x}^{3}}+{{x}^{2}}+x+2=0$. We are going to analyze the factors of the equation and then try to configure each of them equaling to zero. As it is a 3rd degree equation, we are going to find the 3 roots from this given equation, may it be real and complex.

Complete step-by-step answer:
According to the given problem, to start with, we are trying to factorize the equation,
${{x}^{3}}+{{x}^{2}}+x+2=0$
Subtracting 2 from both sides we get,
$\Rightarrow {{x}^{3}}+{{x}^{2}}+x=-2$
Now we will try to transform the equation into terms of monomial, so that we can go ahead with our given problem and get a form of the equation so that the problem can be solved.
$\Rightarrow {{x}^{3}}+{{x}^{2}}+\dfrac{1}{3}x+\dfrac{1}{27}+\dfrac{2}{3}x+\dfrac{2}{9}+\dfrac{47}{27}=0$
After more simplification,
$\Rightarrow {{x}^{3}}+3{{x}^{2}}.\dfrac{1}{3}+3x.\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{2}{3}\left( x+\dfrac{1}{3} \right)+\dfrac{47}{27}=0$
Now, we get, after using the algebraic formulas,
$\Rightarrow {{\left( x+\dfrac{1}{3} \right)}^{3}}+\dfrac{2}{3}\left( x+\dfrac{1}{3} \right)+\dfrac{47}{27}=0$
So, by factorizing, we are getting the equation as a form of,
${{\left( x+\dfrac{1}{3} \right)}^{3}}+\dfrac{2}{3}\left( x+\dfrac{1}{3} \right)+\dfrac{47}{27}=0$
This is not giving us any proper conclusion about what the roots of the equations can be. We can find a solution of the equation which may not be a proper fraction or integer. It might also be a case that the solution we get will be a complex solution. But no proper methods are there to solve these type of problems. That is why we are concluding that none of the known solving methods are giving us any idea about how can we solve this equation.

Hence, the correct option is, (d) None of the above

Note: In relation to quadratic equations, imaginary numbers (and complex numbers) occur when the value under the radical portion of the quadratic formula is negative. When this occurs, the equation has no roots (zeros) in the set of real numbers. The roots belong to the set of complex numbers, and will be called "complex roots" (or "imaginary roots"). These complex roots will be expressed in the form a + bi. The fact which is also known that, ${{i}^{2}}=-1$ .