How do you factor $3{{x}^{2}}-18x-48$?

Hint: First take out ‘3’ from each term of the equation. Then find two numbers whose sum or difference is the remaining coefficient of ‘x’ and multiplication is the result after taking ‘3’ out from the constant term. Split the given equation in terms of multiplication of two factors and do the necessary simplification to obtain the required result.

Complete step by step solution:
Factorization: It is simply the resolution of a polynomial into factors such that when multiplied together they will result in an original polynomial. In the factorization method, we can reduce any algebraic or quadratic equation into its simpler form, where the equations are represented as the product of factors.
The equation we have, $3{{x}^{2}}-18x-48$
Taking out ‘3’ from each term, we get
$\Rightarrow 3\left( {{x}^{2}}-6x-16 \right)$
Now, we have to find two numbers whose sum is ‘$-6$’ and multiplication is ‘$-16$’.
Thus, the numbers are ‘2’ and ‘$-8$’.
Hence, our equation can be written as
$\begin{align}
  & \Rightarrow 3\left( {{x}^{2}}-6x-16 \right) \\
 & \Rightarrow 3\left( {{x}^{2}}+2x-8x-16 \right) \\
\end{align}$
Taking common ‘x’ from first two terms and ‘$-8$’ from last two terms, we get
$\Rightarrow 3\left( x\left( x+2 \right)-8\left( x+2 \right) \right)$
Again taking common $\left( x+2 \right)$ from both the terms, we get
$\Rightarrow 3\left( x+2 \right)\left( x-8 \right)$
This is the required solution of the given question.

Note: As each and every term of the given equation is divisible by ‘3’, so taking common ‘3’ out should be the first approach for the factorization. We can also execute the factorization by completing the square method after taking ‘3’ out.