Question

How do you factor \[{x^3} - 2{x^2} + 4x - 8?\]

Answer 1

Hint:The given question describes the operation of addition/ subtraction/ multiplication/ division.We have to find the value \[x\] from the given equation. To make easy calculations we have to separate \[{x^2}\] terms and constant terms from the given equation. We need to know the basic algebraic formulae to make an easy calculation. Also, we need to know how to take the square root value of negative numbers.

Complete step by step solution:
The given quadratic equation is shown below,
 \[{x^3} - 2{x^2} + 4x - 8 = 0\] \[ \to \left( 1 \right)\]
To find the value \[x\] from the above equation we would try to simplify the above equation.
First, we need to separate \[{x^2}\] from the equation \[\left( 1 \right)\] , we get
 \[\left( 1 \right) \to {x^3} - 2{x^2} + 4x - 8 = 0\]
 \[{x^2}\left( {x - 2} \right) + 4x - 8 = 0 \to \left( 2 \right)\]
Next, we would separate the constant terms from the above equation, so we get
 \[{x^2}\left( {x - 2} \right) + 4\left( {x - 2} \right) = 0 \to \left( 3 \right)\]
In the equation \[\left( 3 \right)\] we can see the term \[\left( {x - 2} \right)\] as a common term, so we get
 \[\left( {x - 2} \right)\left( {{x^2} + 4} \right) = 0\]

So, the value of \[x\] can be found in two cases as follows,
Case: 1
 \[\left( {x - 2} \right) = 0 \\
x = 2 \\
\]
Case: 2
 \[{x^2} + 4 = 0 \\
{x^2} = - 4 \\
x = \sqrt { - 4} \\\]
So, we get
 \[x = \pm 2j\]
(Here \[j\] is the imaginary part)

So, the final answer is,
The value of \[x = 2\] (or) \[x = + 2j\] (or) \[x = - 2j\] .

Note: This question involves the operation of addition/ subtraction/ multiplication/ division. To find the value \[x\] from the given quadratic equation we would try to simplify the equation by separating the common terms.

 Note that if we have a negative number inside the root, we have to put \[j\] when the negative number is taken out from the square root. Remember that when \[j\] is multiplied with \[j\] the answer becomes \[ - 1\] Remember that the value of \[{(j)^2}\] is equal to the value of \[{( - j)^2}\] . That’s why we put \[j\] instead of a symbol inside the root.