How do I find all the rational zeros of a function like $f(x) = {x^3} - 7x - 6$ ?

Hint: The given equation is a polynomial equation of degree 3, for finding the roots of an equation having degree more than 2, we first convert it into the equation of degree by which rest of the roots can be found using any of the methods like factorization, completing the square method, quadratic formula, etc.

Complete step-by-step answer:
In this question, we have to find the rational roots of this equation, so we will accept only rational roots and reject the roots (if any) which do not fulfill this condition.
At x=1,
$
  f(x) = {(1)^3} - 7(1) - 6 \\
  f(x) = 1 - 7 - 6 \\
  f(x) = - 12 \\
   \Rightarrow f(x) \ne 0 \;
 $
So, \[x - 1 = 0\] is not a factor of the given equation.
At x=2,
$
  f(x) = {(2)^3} - 7(2) - 6 \\
  f(x) = 8 - 14 - 6 \\
  f(x) = - 12 \\
   \Rightarrow f(x) \ne 0 \;
 $
So, \[x - 2 = 0\] is not a factor of the given equation.
At x=-1,
$
  f(x) = {( - 1)^3} - 7( - 1) - 6 \\
  f(x) = - 1 + 7 - 6 \\
   \Rightarrow f(x) = 0 \;
 $
So, $x + 1 = 0$ is a solution of the given equation. Using this we can rewrite ${x^3} - 7x - 6$ as –
$
  {x^3} + {x^2} - {x^2} - x - 6x - 6 = 0 \\
   \Rightarrow {x^2}(x + 1) - x(x + 1) - 6(x + 1) \\
   \Rightarrow {x^3} - 7x - 6 = (x + 1)({x^2} - x - 6) \\
   \Rightarrow f(x) = (x + 1)({x^2} - x - 6) \;
 $
So, to find the factors, we get –
${x^2} - x - 6 = 0$
Solving the above equation by factorization –
$
  {x^2} - 3x + 2x - 6 = 0 \\
  x(x - 3) + 2(x - 3) = 0 \\
  (x - 3)(x + 2) = 0 \\
   \Rightarrow x = 3,\,x = - 2 \;
 $
So, $x - 3 = 0$ and $x + 2 = 0$ are the other two factors.
Hence, the factors of the function $f(x) = {x^3} - 7x - 6$ are $x + 1 = 0$ , $x - 3 = 0$ and $x + 2 = 0$ .
So, the correct answer is “$x + 1 = 0$ , $x - 3 = 0$ and $x + 2 = 0$”.

Note: While finding the roots by the hit and trial method, if the answer doesn’t come close to zero after two to three trials then we can skip to bigger values or values off the opposite sign to get closer to zero. After converting the given polynomial equation into a quadratic equation, we find the solution of the equation using factorization, the condition for factorising a quadratic equation of the form $a{x^2} + bx + c = 0$ is that $a \times c = {b_1} \times {b_2}$ and ${b_1} + {b_2} = b$ .