How do you factor \[3{{x}^{2}}-75\]?

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 same 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 but here we do not have b term.so we don’t need to split the b term.

Complete step by step answer:
In the given quadratic polynomial \[3{{x}^{2}}-75\], we can write 75 as a product of 3 and 25.
Then the equation will be
\[\Rightarrow 3{{x}^{2}}-3(25)\]
By taking 3 common from both of terms, we get
\[\Rightarrow 3({{x}^{2}}-25)\]
Now we add \[5x\] and \[-5x\] to \[{{x}^{2}}-25\]. Then the equation will be
\[\Rightarrow 3({{x}^{2}}+5x-5x-25)\]
We take the terms common in the first 2 terms and last 2 terms. Then we get
\[\Rightarrow 3(x(x+5)-5(x+5))\]
Here, we have \[(x+5)\] in common then
\[\Rightarrow 3((x-5)(x+5))\]
\[\therefore 3{{x}^{2}}-75=3((x-5)(x+5))\] is the required answer.

Note:
While solving these types of problems, we need to have enough knowledge on the methods of factoring an expression completely. It is better if we avoid calculation or simplification mistakes. We can solve the above equation easily by using the formula \[{{x}^{2}}-{{y}^{2}}=(x+y)(x-y)\]. If a and c are not perfect squares and b terms cannot be split such that the sum of those parts is b term and product is same as that of product of a and c terms. In such cases, we better go with the quadratic formula.