How do you factor ${x^3} + 7{x^2} + 14x + 8$ ?

Hint: First of all, we will try to simplify the given equation in the question by adding and subtracting $x$ I the equation. Then, use the identity ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$. Solve the remaining quadratic polynomial by taking common from the equation.

Complete Step by Step Solution:
We have to factor the cubic polynomial, ${x^3} + 7{x^2} + 14x + 8$ .
So, first of all, we will try to simplify the given equation in the question. Therefore, to make it easy add and subtract $x$ in the equation –
$
   \Rightarrow {x^3} - x + 7{x^2} + 14x + x + 8 \\
   \Rightarrow {x^3} - x + 7{x^2} + 15x + 8 \\
 $
Now, we will take $x$ common from ${x^3} - x$ in the above equation, we get –
$ \Rightarrow x\left( {{x^2} - 1} \right) + 7{x^2} + 15x + 8 \cdots \left( 1 \right)$
We know that, the identity of the difference in the squares is –
${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$
Now, applying the identity of the difference in the squares in the equation (1)
So, $\left( {{x^2} - {1^2}} \right)$ can be written as $\left( {x - 1} \right)\left( {x + 1} \right)$ and putting this in the equation (1), we get –
$ \Rightarrow x\left( {x - 1} \right)\left( {x + 1} \right) + 7{x^2} + 15x + 8 \cdots \left( 2 \right)$
Taking $7{x^2} + 15x + 8$ part and solving it form the above equation –
$
   \Rightarrow 7{x^2} + 15x + 8 = 7{x^2} + 7x + 8x + 8 \\
   \Rightarrow 7{x^2} + 15x + 8 = 7x\left( {x + 1} \right) + 8\left( {x + 1} \right) \\
 $
Putting this value of $7{x^2} + 15x + 8$ in the equation (2), we get –
$
   \Rightarrow x\left( {x + 1} \right)\left( {x - 1} \right) + \left[ {7x\left( {x + 1} \right) + 8\left( {x + 1} \right)} \right] \\
   \Rightarrow \left( {x + 1} \right)\left( {x - 1} \right) + \left( {x + 1} \right)\left( {7x + 8} \right) \\
 $
Now, taking $\left( {x + 1} \right)$ common from the above whole equation, we get –
$
   \Rightarrow \left( {x + 1} \right)\left[ {x\left( {x - 1} \right)\left( {7x + 8} \right)} \right] \\
   \Rightarrow \left( {x + 1} \right)\left[ {{x^2} - x + 7x + 8} \right] \\
   \Rightarrow \left( {x + 1} \right)\left[ {{x^2} + 6x + 8} \right] \cdots \left( 3 \right) \\
 $
Now, again taking $\left[ {{x^2} + 6x + 8} \right]$ part and solving it to get the factors –
$
   \Rightarrow {x^2} + 6x + 8 = {x^2} + 4x + 2x + 8 \\
   \Rightarrow {x^2} + 6x + 8 = x\left( {x + 4} \right) + 2\left( {x + 4} \right) \\
 $
Putting this value of $\left( {{x^2} + 6x + 8} \right)$ in equation (3), we get –
$ \Rightarrow \left( {x + 1} \right)\left[ {x\left( {x + 4} \right) + 2\left( {x + 4} \right)} \right]$
We can also write the above equation in the following way –
$ \Rightarrow \left( {x + 1} \right)\left( {x + 4} \right)\left( {x + 2} \right)$

Hence, this is the required factorization of the equation ${x^3} + 7{x^2} + 14x + 8$.

Note:
The polynomial equation is the algebraic expression that contains alphabetical as well as numerical values but when the alphabets representing an unknown variable quantity are raised to some power such that the exponent is non – negative integer then, the algebraic expression becomes a polynomial equation.
Many students can make mistakes and just leave the question during the finding of factors for quadratic polynomials but they should solve both the quadratic polynomials we got during the solving of the question.