How do you factor the expression ${x^2} + x - 42?$

Hint: Factorizing reduces the higher degree equation into its linear equation. In the above given question, we need to reduce the quadratic equation into its simplest form in such a way that addition of products of the factors of first and last term should be equal to the middle term i.e. $x$.

Complete step by step answer:
The equation $ a{x^2} + bx + c$ is a general way of writing quadratic equations where a, b and c are numbers.
In the above expression,
a = $1$, b = $1$, c = $ - 42$
${x^2} + x - 42$
First step is by multiplying the coefficient of ${x^2}$ and the constant term -42, we get $ - 42{x^2}$.
After this, factors of $ - 42{x^2}$ should be calculated in such a way that their addition should be equal to x.
Factors of 42 can be 7 and 6.
Since, we consider a negative sign of 42 one of the factors should be negative so that multiplication result is negative and by adding these two factors we should get positive 1$x$.
Therefore, we can add in the following way $7{x^{}}$+ ($ - 6x$) = $x$.
So, further we write the equation by equating it with zero and splitting the middle term according to the factors.
$
   \Rightarrow {x^2} + x - 42 = 0 \\
   \Rightarrow {x^2} + 7x - 6x - 42 = 0 \\
 $
Now, by grouping the first two and last two terms we get common factors.
$
   \Rightarrow x(x + 7) - 6(x + 7) = 0 \\
   \Rightarrow (x - 6)(x + 7) = 0 \\
 $
Taking x common from the first group and -6 common from the second we get the above equation.
Therefore, by solving the above quadratic equation we get factors 6 and -7.

Note: In quadratic equation, an alternative way of finding the factors is by using a formula which is given below:
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
By substituting the values of a=1, b=1 and c=-42 we get the factors of x.
$
  x = \dfrac{{ - 1 \pm \sqrt {{{(1)}^2} - 4(1)( - 42)} }}{{2(1)}} \\
    \\
 $
$x = $6 or $x = $-7