How do you factor \[{x^2} + 4x - 12\]?

Hint: Here in this question, we have to find the factors, the given equation is in the form of a quadratic equation. This is a quadratic equation for the variable x. By using the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\], we can determine the roots of the equation and factors are given by (x – root1) (x – root 2).

Complete step-by-step solution:
The question involves the quadratic equation. To the quadratic equation we can find the roots by factoring or by using the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]. So the equation is written as \[{x^2} + 4x - 12\].
In general, the quadratic equation is represented as \[a{x^2} + bx + c = 0\], when we compare the above equation to the general form of equation the values are as follows. a=1 b=4 and c=-12. Now substituting these values to the formula for obtaining the roots we have
\[roots = \dfrac{{ - 4 \pm \sqrt {{{(4)}^2} - 4(1)( - 12)} }}{{2(1)}}\]
On simplifying the terms, we have
\[ \Rightarrow roots = \dfrac{{ - 4 \pm \sqrt {16 + 48} }}{2}\]
Now add 16 to 48 we get
\[ \Rightarrow roots = \dfrac{{ - 4 \pm \sqrt {64} }}{2}\]
The number 64 is a perfect square so we can take out from square root we have
\[ \Rightarrow roots = \dfrac{{ - 4 \pm 8}}{2}\]
Therefore, we have \[root1 = \dfrac{{ - 4 + 8}}{2} = \dfrac{4}{2} = 2\] or \[root2 = \dfrac{{ - 4 - 8}}{2} = \dfrac{{ - 12}}{2} = - 6\].
The roots for the quadratic equation when we find the roots by using formula is given by (x – root1) (x – root 2)
Substituting the roots values, we have
\[ \Rightarrow \left( {x - 2} \right)\left( {x - ( - 6)} \right)\]
On simplifying we have
\[ \Rightarrow (x - 2)(x + 6)\]

Hence, we have found the factors for the given equation

Note: To find the factors for algebraic equations or expressions, it depends on the degree of the equation. If the equation contains a square then we have two factors. If the equation contains a cube then we have three factors. Here this equation also contains 3 factors, the two factors are imaginary.