Question

How do you factor \[{x^3} + 6{x^2} + 12x + 8?\]

Answer 1

Hint: This question involves the operation of addition/ subtraction/ multiplication/ division. We need to know the basic algebraic formulae to make an easy calculation. We need to know how to find a common factor from the given equation. We need to know the basic conditions with the involvement of power terms.

Complete step-by-step answer:
The given equation is shown below,
 \[{x^3} + 6{x^2} + 12x + 8 \to \left( 1 \right)\]
To solve the above equation we need to find a common factor from the above equation. In the equation \[\left( 1 \right)\] , we have found that, \[6x\] is common in the term \[6{x^2}\] and \[12x\] . So, the equation \[\left( 1 \right)\] can also be written as,
 \[
  \left( 1 \right) \to {x^3} + 6{x^2} + 12x + 8 \\
   = {x^3} + 6x\left( {x + 2} \right) + 8 \;
 \]
The above equation can also be written as,
 \[ \Rightarrow {x^3} + 8 + 6x\left( {x + 2} \right) \to \left( 2 \right)\]
We know that \[8\] can be written as \[{2^3}\] . So, the equation \[\left( 2 \right)\] becomes,
 \[ \Rightarrow {x^3} + {2^3} + 6x\left( {x + 2} \right) \to \left( 3 \right)\]
We know that,
 \[{a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)\]
By using the above-mentioned formula \[\left( {{x^3} + {2^3}} \right)\] can be written as
 \[\left( {{x^3} + {2^3}} \right) = \left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right) \to \left( 4 \right)\]
Let’ substitute the equation \[\left( 4 \right)\] in the equation \[\left( 3 \right)\] , we get
 \[{x^3} + {2^3} + 6x\left( {x + 2} \right) = \left( {{x^3} + {2^3}} \right) = \left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right) + 6x\left( {x + 2} \right) \to \left( 5 \right)\]
In the above equation, the term \[\left( {x + 2} \right)\] is common in both terms. So, the equation \[\left( 5 \right)\] becomes,
 \[ \Rightarrow \left( {x + 2} \right)\left( {{x^2} - 2x + 4 + 6x} \right)\]
By simplifying the above equation \[\left( {6x - 2x = 4x} \right)\] , we get
 \[ \Rightarrow \left( {x + 2} \right)\left( {{x^2} + 4x + 4} \right)\]
 \[4\] Can also be written as \[{2^2}\] and \[4x\] can be written as \[2 \times 2 \times x\] , so we get
 \[ \Rightarrow \left( {x + 2} \right)\left( {{x^2} + 2 \times 2 \times x + {2^2}} \right) \to \left( 6 \right)\]
We know that,
 \[{a^2} + 2ab + {b^2} = {\left( {a + b} \right)^2}\]
By using the above formula, we can write,
 \[\left( {{x^2} + 2 \times 2 \times x + {2^2}} \right) = {\left( {x + 2} \right)^2} \to \left( 7 \right)\]
(Here \[a = x\] and \[b = 2\] )
By substituting the equation \[\left( 7 \right)\] in the equation \[\left( 6 \right)\] , we get
 \[ \Rightarrow \left( {x + 2} \right)\left( {{x^2} + 2 \times 2 \times x + {2^2}} \right) = \left( {x + 2} \right){\left( {x + 2} \right)^2}\]
We know that,
 \[a \cdot {a^2} = {a^3}\]
So, we get
 \[\left( {x + 2} \right){\left( {x + 2} \right)^2} = {\left( {x + 2} \right)^3}\]
So, the final answer is,
 \[{x^3} + 6{x^2} + 12x + 8 = {\left( {x + 2} \right)^3}\]
So, the correct answer is “ \[{x^3} + 6{x^2} + 12x + 8 = {\left( {x + 2} \right)^3}\] ”.

Note: This question describes the operation of addition/ subtraction/ multiplication/ division. Remember the basic algebraic formulae to solve these types of questions. Also, note that we won’t take \[1\] it as a common factor. Note that anything power zero will be one and zero power anything will be zero. Note that if \[{x^3}\] is present in the given equation we would find three values for \[x\] .