How do you factor \[12{{x}^{2}}-5x-2\]?

Hint: When we have a polynomial of the form \[a{{x}^{2}}+bx+c\], we can factor this quadratic with splitting up the b term into two terms based on the signs of a and c terms. If we have the opposite sign for both a and c terms, then we split the b term into two parts. These parts are formed such that the sum of them is b term itself and their product is the same as that of the product of a and c terms.

Complete step by step answer:
In the given quadratic polynomial \[12{{x}^{2}}-5x-2\], the coefficient of \[{{x}^{2}}\] and constant terms are of the opposite sign and their product is -8. That is we have a and c coefficients with opposite signs in the polynomial of the form \[a{{x}^{2}}+bx+c\].
Hence, we would split -5, which is the coefficient of x, in two parts, whose sum is -5 and product is -24. These are -8 and 3.
So, we write it as
\[\Rightarrow 12{{x}^{2}}-5x-2\]
We now split \[-5x\] into \[-8x\] and \[3x\].
\[\Rightarrow 12{{x}^{2}}-8x+3x-2\]
We take the terms common in first 2 terms and last 2 terms
\[\Rightarrow 4x(3x-2)+1(3x-2)\]
Here, we have \[(3x-2)\] in common then
\[\Rightarrow (4x+1)(3x-2)\]

\[\therefore (4x+1)(3x-2)\] is the required answer.

Note: If the coefficient of \[{{x}^{2}}\] and constant terms are the same, then we factor the polynomial by splitting up the b term into two terms. Here also we have to split b term such that sum of those parts is b term and product is same as that of product of a and c terms. Factoring by grouping will not always work. In such cases we better go with the quadratic formula.