Question

How do you factor $y = {x^3} - 2{x^2} + x - 2$?

Answer 1

Hint: As the given equation is cubic in one variable. First, make the pairs of two terms. After that, make common in both pairs. After that take commons from both pairs. Then, substitute ${i^2}$ in place of $ - 1$. Then, again factor the terms. So, the factored terms are the desired result.

Complete step-by-step answer:
In mathematics, the number system is the branch that deals with various types of numbers possible to form and easy to operate with different operators such as addition, multiplication, and so on. Another system for variables in mathematics is an algebra system in which we have equations corresponding to some variables through which we can evaluate the values of those variables.
Here, we have one equation to factorize which is involving variable x. Factors are those numbers that completely divide the given number without leaving any remainder. Similarly, factors of an equation will completely divide the equation without leaving any remainder.
We have been given an equation, $y = {x^3} - 2{x^2} + x - 2$.
We have to find the factors of the given equation.
First, make pairs of two terms,
$ \Rightarrow y = \left( {{x^3} - 2{x^2}} \right) + \left( {x - 2} \right)$
Take common from both pairs,
$ \Rightarrow y = {x^2}\left( {x - 2} \right) + \left( {x - 2} \right)$
Again, take commonly from the terms,
$ \Rightarrow y = \left( {x - 2} \right)\left( {{x^2} + 1} \right)$
As we know that
${i^2} = - 1$
Substitute $ - {i^2}$ in place of 1,
$ \Rightarrow y = \left( {x - 2} \right)\left( {{x^2} - {i^2}} \right)$
We know that, $\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)$, we get
$ \Rightarrow y = \left( {x - 2} \right)\left( {x + i} \right)\left( {x - i} \right)$

Hence, the factors of the equation $y = {x^3} - 2{x^2} + x - 2$ is $y = \left( {x - 2} \right)\left( {x + i} \right)\left( {x - i} \right)$.

Note:
The key step for solving this problem is the knowledge of the algebraic system of equations. To solve any particular algebraic equation, we require the same number of equations as there are several variables present. Since we have one equation and a single variable so we can factorize the problem.