
Factorize the following expression:
${x^3} - 2{x^2} - x + 2$.
Answer
604.8k+ views
Hint – In order to solve this problem we have to write each factor of this expression raised to the power of one. We need to form a group of terms which have a common factor. Using this we can factorize the given expression.
Complete step-by-step answer:
The given expression is ${x^3} - 2{x^2} - x + 2$.
We can clearly see that we can take ${x}^2$ common from the first two terms of the expression.
$ \Rightarrow ({x^2}) (x - 2)- x - 2 \\$
We can clearly see that we can take $x - 2$ common from the two terms of the expression. The equation can be written as:
$ \Rightarrow({x^2}) (x - 2) - (x - 2)$
On taking $x - 2$ common from the equation we get the equation as:
$
\Rightarrow (x - 2)({x^2}) - (x - 2)1 \\
\Rightarrow (x - 2)({x^2} - 1) \\
$
We know that $({a^2} - {b^2}) = (a - b)(a + b)$
So, we can write $ \Rightarrow (x - 2)({x^2} - 1) = (x - 2)(x - 1)(x + 1)$
Hence the answer is $(x - 2)(x - 1)(x + 1)$.
Note – In these types of problems of cubic equations you just need to take something common so that all terms are simplified to the terms raised to the power one then we can say that the equation is totally factorized. Proceeding like this will solve your problem.
Complete step-by-step answer:
The given expression is ${x^3} - 2{x^2} - x + 2$.
We can clearly see that we can take ${x}^2$ common from the first two terms of the expression.
$ \Rightarrow ({x^2}) (x - 2)- x - 2 \\$
We can clearly see that we can take $x - 2$ common from the two terms of the expression. The equation can be written as:
$ \Rightarrow({x^2}) (x - 2) - (x - 2)$
On taking $x - 2$ common from the equation we get the equation as:
$
\Rightarrow (x - 2)({x^2}) - (x - 2)1 \\
\Rightarrow (x - 2)({x^2} - 1) \\
$
We know that $({a^2} - {b^2}) = (a - b)(a + b)$
So, we can write $ \Rightarrow (x - 2)({x^2} - 1) = (x - 2)(x - 1)(x + 1)$
Hence the answer is $(x - 2)(x - 1)(x + 1)$.
Note – In these types of problems of cubic equations you just need to take something common so that all terms are simplified to the terms raised to the power one then we can say that the equation is totally factorized. Proceeding like this will solve your problem.
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