How do you factorize the expression $12{x^2} - 75?$

Hint: As we know that factorising is the reverse of expanding brackets, it is an important way of solving equations. The first step of factoring an expression is to take out any common factors which the terms have. So if we were asked to factor the expression ${x^2} + x$, since $x$ goes into both terms, we would write $x(x + 1)$. Here we will use identities which will help us to factorise an algebraic expression easily i.e. ${(a - b)^2} = {a^2} - 2ab + {b^2}$ and ${a^2} - {b^2} = (a + b)(a - b)$.

Complete step by step answer:
Here we will use some identities to help like the difference of square identity:
${a^2} - {b^2} = (a + b)(a - b)$, but looking at the expression we can get the idea that $12$ and $75$ are not perfect squares. So we have to see if there is any common factor between both the terms and if there is any, then will take out the common factor among the two terms.
In the above given equation, $3$ is a common factor to both the terms, so we'll take it out: $12{x^2} - 75 = 3(4{x^2} - 25)$. Now the expressions in the brackets have $4$ and $25$. Both of the numbers are perfect squares and then we can now apply the identity ${a^2} - {b^2} = (a + b)(a - b)$. So we have $3 \times \{ {(2x)^2} - {5^2}\} \Rightarrow 3(2x + 5)(2x - 5)$ i.e. $a = 2x,b = 5$. It cannot be factored further.
Hence the factored form of the expression is $12{x^2} - 75 = 3(2x + 5)(2x - 5)$.

Note: We should keep in mind while solving these expressions that we use correct identities to factorise the given algebraic expressions and keep checking the negative and positive sign otherwise it will give the wrong answer . Also we should know that the difference of square formula is the identity that is used in the above solution. These are some of the standard algebraic identities. This is as far we can go with real coefficients as the remaining quadratic factors all have complex zeros.