Question

The roots of the equation ${x^3} - 3x - 2 = 0$ are:
(A) $ - 1, - 1,2$
(B) $ - 1,1, - 2$
(C) $ - 1,2, - 3$
(D) $ - 1, - 1, - 2$

Answer 1

Hint: In the given problem, we have to factorise the given polynomial equation and find its roots. The given equation is of degree $3$ and is thus called a cubic equation. For finding the factors of a cubic polynomial, we first find out the root of the polynomial by hit and trial method. Hence, we get one factor using hit and trial and then find the remaining two roots by using various methods like splitting the middle term and using quadratic formula.

Complete step-by-step solution:
So, we have, ${x^3} - 3x - 2 = 0$
Now, let us consider the given polynomial ${x^3} - 3x - 2$ as $p\left( x \right)$.
Now, using hit and trial method,
Putting $x = - 1$in the polynomial $p\left( x \right)$,
\[p(x = - 1) = {\left( { - 1} \right)^3} - 3\left( { - 1} \right) - 2\]
Simplifying the calculations, we get,
$ \Rightarrow $\[p(x = - 1) = \left( { - 1} \right) + 3 - 2\]
$ \Rightarrow $$p\left( {x = - 1} \right) = 0$
Hence, $p( - 1) = 0$
So, we conclude that $x = - 1$is a root of the polynomial. Hence, by factor theorem, $\left( {x + 1} \right)$ is a factor of the polynomial.
So, we try to factor out the factor $\left( {x + 1} \right)$ from the equation given. Hence, we get,
${x^3} - 3x - 2 = 0$
\[ \Rightarrow {x^2}\left( {x + 1} \right) - x\left( {x + 1} \right) - 2\left( {x + 1} \right) = 0\]
Taking out the common term, we get,
\[ \Rightarrow \left( {{x^2} - x - 2} \right)\left( {x + 1} \right) = 0\]
Now, $p(x) = \left( {{x^3} - 3x - 2} \right) = \left( {x + 1} \right)\left( {{x^2} - x - 2} \right)$
Now quadratic polynomials $\left( {{x^2} - x - 2} \right)$ can be factored further by making use of splitting the middle term method.
For factoring the quadratic polynomial $\left( {{x^2} - x - 2} \right)$ , we can use the splitting method in which the middle term is split into two terms such that the sum of the terms gives us the original middle term and product of the terms gives us the product of the constant term and coefficient of ${x^2}$.
So, $\left( {{x^2} - x - 2} \right)$
$ \Rightarrow {x^2} - 2x + x - 2$
We split the middle term $ - x$ into two terms $ - 2x$ and $x$ since the product of these terms, $ - 2{x^2}$ is equal to the product of the constant term and coefficient of ${x^2}$ and sum of these terms gives us the original middle term, $ - x$.
$ \Rightarrow x\left( {x - 2} \right) + \left( {x - 2} \right)$
Taking the common factor out, we get,
$ \Rightarrow \left( {x + 1} \right)\left( {x - 2} \right)$
Therefore, we get the equation as,
$ \Rightarrow \left( {x + 1} \right)\left( {x - 2} \right)\left( {x + 1} \right) = 0$
Now, we know that any one of the brackets has to be zero if the product of all the three factors is equal to zero. So, we get,
$\left( {x + 1} \right) = 0$, $\left( {x - 2} \right) = 0$ and $\left( {x + 1} \right) = 0$
Now, we can solve for the values of x by using the method of transposition in the above equations.
So, we get,
$ \Rightarrow x = - 1$, $x = 2$ and $x = - 1$
So, the roots of the equation ${x^3} - 3x - 2 = 0$ are $ - 1$, $ - 1$ and $2$.
Hence, option (A) is the correct answer.

Note: Such cubic equations can be solved by firstly using the hit and trial method to find one zero and then using the splitting the middle term and factorisation to solve for the remaining roots. We must have a strong grip over the concepts of algebraic expressions and simplification rules. Also, we should take care of calculations while solving for the roots of the equation.