Question

How do you factorize $3{x^2} - 12$ ?

Answer 1

Hint: Here, we have to factorize $3{x^2} - 12$. By applying a known mathematical formula ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$, we can factorize the given algebraic expression.

Complete step by step answer:
Given: we have to factorize $3{x^2} - 12$.
Here, it is clearly visible that $3$ is common in both the terms. So, first of call put $3$ out of both terms and then we get,
$ = 3\left( {{x^2} - 4} \right)$
We can write $3\left( {{x^2} - 4} \right)$ as $3\left( {{x^2} - {2^2}} \right)$.
Now, by applying above given formula ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$ we can write
$ = 3\left( {x + 2} \right)\left( {x - 2} \right)$

Thus, we can factorize $3{x^2} - 12$ as $3\left( {x + 2} \right)\left( {x - 2} \right)$.

Note: Factorization of a quadratic polynomial:
Suppose $a{x^2} + bx + c$ is a quadratic polynomial. If we have to factorise this quadratic polynomial then we have to break the coefficient of $x$ that is $b$ into two parts such that their product is equal to the product of the coefficient of ${x^2}$ and the constant term. Then take common from the two terms and then take common from the last two terms.
eg:- the given quadratic polynomial is ${x^2} + 5x + 6$.
Firstly, we have to break the coefficient of $x$ that is $5$ into two parts that is $2$ and $3$ because their product is equal to the product of the coefficient of ${x^2}$ and the constant term.
So, we can write ${x^2} + 5x + 6$ as ${x^2} + \left( {2 + 3} \right)x + 6$
$ = {x^2} + 2x + 3x + 6$
Now, take $x$ as common from the first two terms and take $3$ as common from the last two terms and then we get
$ = x\left( {x + 2} \right) + 3\left( {x + 2} \right)$
It is clearly visible that $\left( {x + 2} \right)$ is common in both terms. So, taking $\left( {x + 2} \right)$ as common we can write
$ = \left( {x + 2} \right)\left( {x + 3} \right)$
Alternatively, the given polynomial $3{x^2} - 12$ can be factorized by this method. Firstly, take $3$ as common from both the terms and then factorize ${x^2} - 4$. To factorize ${x^2} - 4$ we have to add and subtract $2x$ and then take $x$ as common from the first two terms and $2$ as common from the last two terms and follow the steps as shown above.