Question

Find the zeroes of the polynomial ${x^3} + 2{x^2} - x - 2$.
$
  {\text{A}}{\text{. }}1, - 1, - 2 \\
  {\text{B}}{\text{. }}1,3,7 \\
  {\text{C}}{\text{. }}1, - 2,4 \\ $
${\text{D}}{\text{. }}$None of the above

Answer 1

Hint: Here, we will proceed by equating the given cubic polynomial to zero in order to get the cubic equation which will be factorized so that the given polynomial ${x^3} + 2{x^2} - x - 2$ can be represented as the product of its factors.

Complete step-by-step answer:
The given cubic polynomial is ${x^3} + 2{x^2} - x - 2$
In order to find the roots or zeros of the given cubic polynomial, we will equate this polynomial with zero and this equation is the cubic equation corresponding to the given cubic polynomial.
The cubic equation is ${x^3} + 2{x^2} - x - 2 = 0$
By taking ${x^2}$ and 1 common from the first two terms and the last two terms respectively, we get
$ \Rightarrow {x^2}\left( {x + 2} \right) - 1\left( {x + 2} \right) = 0$
By taking $\left( {x + 2} \right)$ common from the two terms given in the above equation, we get
$ \Rightarrow \left( {x + 2} \right)\left( {{x^2} - 1} \right) = 0$
Either $
  \left( {x + 2} \right) = 0 \\
   \Rightarrow x = - 2 \\
 $ or $
  \left( {{x^2} - 1} \right) = 0 \\
   \Rightarrow {x^2} = 1 \\
   \Rightarrow x = \pm \sqrt 1 \\
   \Rightarrow x = \pm 1 \\
 $
So, the zeros or roots of the polynomial ${x^3} + 2{x^2} - x - 2$ are x=1,-1,-2.
Hence, option A is correct.

Note: In this particular problem, the cubic equation obtained is easily factored by taking something common from the terms but there are most of the cubic equations, we cannot be solved by this approach. In order to solve those cubic equations, we apply a hit and trial approach so that the LHS of the cubic equation for the root or zero of that equation comes out being equal to the RHS.