Question

How do you solve ${x^3} - 2{x^2} - x + 2 = 0$?

Answer 1

Hint: In the question, we are provided with a cubic equation. So, we will get three values(roots) for the given equation. Firstly, determining the first root by hit and trial method and then the rest two roots by the same methods for solving any quadratic equation.

Complete step by step answer:
We started finding the roots of the equation given in the question by hit and trial method. Checking for different values of $x$ as
$x = 1, - 1,2, - 2$ by putting such values in the
given equation and the value which satisfies the equation will be our first root this is the Hit and trial method.
Taking $x = 1$, putting the value in the above expression
${\left( 1 \right)^3} - 2{\left( 1 \right)^2} - 1 + 2 = 1 - 2 - 1 + 2 = 0$
Which is true, hence $x = 1$ satisfies the given expression. So, the first root of the cubic expression. So, the first root of the cubic equation is 1.
Now, we imply that $x - 1$ is a factor of ${x^3} - 2{x^2} - x + 2 = 0$.
Hence, dividing ${x^3} - 2{x^2} - x + 2$ by $\left( {x - 1} \right)$, which gives ${x^2} - x - 2$.
We are left with a quadratic equation. To find the remaining two roots, we are using the split the middle term method.
${x^2} - x - 2 = 0$ writing the middle term in terms of $x, - 2x$.
${x^2} + x - 2x - 2 = 0$
Now, taking the common
$x\left( {x + 1} \right) - 2\left( {x + 1} \right) = 0$
$ \Rightarrow \left( {x + 1} \right)\left( {x - 2} \right) = 0$
$ \Rightarrow x = - 1,2$

Hence, we find our three roots which are $x = 1, - 1,2$ which is our required answer.

Note: Number of zeroes (solution) of an expression or an equation depends on the highest degree of the variable. For example, in the above question expression we had the highest power of $x$ to be three.
Hence, we had exactly three solutions. And yes, ZERO can also be a solution to any equation.