How do you factor\[{{x}^{3}}-2{{x}^{2}}+x-2\]?

Hint: In this type of question, during the solving process, we should have a better knowledge of the factorization topic of the algebra chapter. For finding the factor of\[{{x}^{3}}-2{{x}^{2}}+x-2\], first, we will group the first 2 terms and second 2 terms individually. Further, we will do the rest to find the solution. And, also we will see the alternate method for this question.

Complete step by step answer:
Let us solve this question.
In the question, it is asked to factor the equation\[{{x}^{3}}-2{{x}^{2}}+x-2\].
\[{{x}^{3}}-2{{x}^{2}}+x-2=x\times x\times x-2\times x\times x+x-2\]
As we can see from the above equation that \[x\times x={{x}^{2}}\] is common in the first two terms.
So, we can write the above equation as:
\[\Rightarrow {{x}^{3}}-2{{x}^{2}}+x-2={{x}^{2}}(x-2)+(x-2)\]
Now, we can see that a common factor (x-2) is there in the equation.
\[\Rightarrow {{x}^{3}}-2{{x}^{2}}+x-2={{x}^{2}}(x-2)+(x-2)={{x}^{2}}(x-2)+1(x-2)\]
\[\Rightarrow {{x}^{3}}-2{{x}^{2}}+x-2=({{x}^{2}}+1)(x-2)\]
Hence, the factor of \[{{x}^{3}}-2{{x}^{2}}+x-2\] is\[({{x}^{2}}+1)(x-2)\].

Note:
We have an alternate method to solve this question. For factorizing cubic polynomials, the method can be used is in the following:
Before solving this question by an alternate method, let us remember one thing.
If an equation \[{{x}^{n}}\times {{x}^{n-1}}\times {{x}^{n-2}}\times .........\times {{x}^{2}}\times {{x}^{1}}\times 1\] has a root\[{{x}_{1}}\]. Then, \[\left( x-{{x}_{1}} \right)\] is a factor of\[{{x}^{n}}\times {{x}^{n-1}}\times {{x}^{n-2}}\times .........\times {{x}^{2}}\times {{x}^{1}}\times 1\]. And, at\[x={{x}_{1}}\], the value of the equation \[{{x}^{n}}\times {{x}^{n-1}}\times {{x}^{n-2}}\times .........\times {{x}^{2}}\times {{x}^{1}}\times 1\] will be zero.
So, first, we will check if there is any root of the equation\[{{x}^{3}}-2{{x}^{2}}+x-2\].
We find one factor of the equation by trial and error method.
At x=0, the value of \[{{x}^{3}}-2{{x}^{2}}+x-2\] will be -2. So, 0 is not the root of\[{{x}^{3}}-2{{x}^{2}}+x-2\].
At x=1, the value of \[{{x}^{3}}-2{{x}^{2}}+x-2\] will be -2. So, 1 is not the root of\[{{x}^{3}}-2{{x}^{2}}+x-2\].
At x=2, the value of \[{{x}^{3}}-2{{x}^{2}}+x-2\] will be 0.
So, 2 is the root of\[{{x}^{3}}-2{{x}^{2}}+x-2\].
When replacing x by 2 in the equation\[{{x}^{3}}-2{{x}^{2}}+x-2\],
\[\Rightarrow {{2}^{3}}-2\times {{2}^{2}}+2-2\]
\[\Rightarrow 8-8+2-2\]
\[\Rightarrow 0\]
Hence, from here, we get that 2 is the root of the equation\[{{x}^{3}}-2{{x}^{2}}+x-2\].
Therefore, (x-2) is the factor of the equation\[{{x}^{3}}-2{{x}^{2}}+x-2\].
Then by long division method, we will divide \[(x-2)\] by \[{{x}^{3}}-2{{x}^{2}}+x-2\].
\[\dfrac{{{x}^{3}}-2{{x}^{2}}+x-2}{x-2}=\dfrac{{{x}^{3}}-2{{x}^{2}}}{x-2}+\dfrac{x-2}{x-2}\]
Which is also can be written as
\[\Rightarrow \dfrac{{{x}^{3}}-2{{x}^{2}}+x-2}{x-2}=\dfrac{{{x}^{2}}\left( x-2 \right)}{x-2}+\dfrac{x-2}{x-2}\]
\[\Rightarrow \dfrac{{{x}^{3}}-2{{x}^{2}}+x-2}{x-2}={{x}^{2}}+1\]
Hence, from here also, we get that \[\left( {{x}^{2}}+1 \right)\] and \[\left( x-2 \right)\] are the factors of\[{{x}^{3}}-2{{x}^{2}}+x-2\].