Answer
\[{{x}^{2}}-x-12=\left( x+3 \right)\left( x-4 \right)\]
Work Step by Step
Note that this trinomial has leading coefficient is 1, therefore, to factor it, follow the steps below:
Step 1:
Find first two terms whose product is\[{{x}^{2}}\]:
\[{{x}^{2}}-x-12=\left( x\text{ } \right)\left( x\text{ } \right)\]
Step 2:
Find the last two terms that when multiplied gives \[-12\]:
\[\text{Factors of }-12:\text{ }-1,12\text{ }1,-12\text{ }-3,4\text{ }-\text{3},4\text{ 6,}-2\text{ }-2,6\]
Step 3:
Now try various combinations of these factors and look for the pair of factors whose sum is the coefficient of the middle term. That is, \[-1\]in this case.
\[\begin{align}
& \text{Factors of }-12:\text{ }-1,12\text{ }1,-12\text{ }-3,4\text{ 3},-4\text{ 6,}-2\text{ }2,-6 \\
& \text{Sum of factors : }11\text{ }-11\text{ 1 }-1\text{ }4\text{ }-4\text{ } \\
\end{align}\]
Thus from above \[\text{3},-4\] is the required factor of \[-12\] that gives the desired sum.
So that,
\[{{x}^{2}}-x-12=\left( x+3 \right)\left( x-4 \right)\].
