How do you factor \[{{x}^{2}}-x-42\]?

Hint: In this problem, we are going to factorize the given equation. We should know that we have many methods to factorize a quadratic equation. Here, we can take a simple factorization method to factorize the given equation. We can take the constant term -42, if we multiply -7 and 6, the answer is the constant -42 and adding those two we get -1, which is the coefficient of x. Therefore, we can find the factor.

Complete step by step answer:
We know that the given quadratic equation is,
\[{{x}^{2}}-x-42\] ……. (1)
Now, we can factorise the given equation.
We can take the constant term -42, if we multiply -7 and 6, the answer is the constant -42 and adding those two we get -1, which is the coefficient of x.
\[\begin{align}
  & \Rightarrow -7\times 6=-42 \\
 & \Rightarrow -7+6=-1\left( \text{coefficient of x} \right) \\
\end{align}\]
Now we can write the equation (1) as,
\[{{x}^{2}}-7x+6x-42\] ……. (2)
Now we can split the above quadratic equation (2).
We have four terms in the above equation, we can take the common x outside from the first two terms and we can take the common number 6 from the last two terms.
We get,
\[\Rightarrow x\left( x-7 \right)+6\left( x-7 \right)\]
Now, we can take the common term from the above expression, we get
\[\Rightarrow \left( x+6 \right)\left( x-7 \right)\]

Therefore, the factors of the equation \[{{x}^{2}}-x-42\] is \[\left( x+6 \right)\left( x-7 \right)\].

Note: Students make mistakes while finding the equation (2) part, we should know that the factors we get, comes from multiplying two numbers which gives the exact constant term and adding two numbers which gives the coefficient of x for the same equation.