Question

Factorise the polynomial $\left( {4{x^2} - 5} \right)$?

Answer 1

Hint: Given polynomial is of degree 2. Polynomials of degree 2 are known as Quadratic polynomials. Quadratic polynomials can be factored by the help of splitting the middle term method. In this method, the middle term is split into two terms in such a way that the polynomial remains unchanged. But the given polynomial does not have a middle term. Hence, the given polynomial should be factored using algebraic identities.

Complete step by step solution:
Given question requires us to factorise the polynomial $\left( {4{x^2} - 5} \right)$
So, let the polynomial $\left( {4{x^2} - 5} \right)$ be $p(x)$.
Then, $p(x) = (4{x^2} - 5)$
The given polynomial does not have any middle term. Hence, the given polynomial should be factorised using algebraic identities. We can use algebraic identity $\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)$ to factorise the given polynomial.
In order to use the algebraic identity, we should convert the given polynomial into a form that would resemble the algebraic identity.
$p(x) = {(2x)^2} - {\left( {\sqrt 5 } \right)^2}$
Now, using identity, \[\left( {{a^2} - {b^2}} \right) = (a + b)(a - b)\], we get,
\[p(x) = (2x + \sqrt 5 )(2x - \sqrt 5 )\]
So, the factorized form of the given polynomial is \[(2x + \sqrt 5 )(2x - \sqrt 5 )\].

Note: Similar to quadratic polynomials, quadratic equations can also be solved using factorisation method. Besides factorisation, there are various methods to solve quadratic equations such as completing the square method, splitting the middle term and using the Quadratic formula. Splitting of the middle term can be a tedious process at times when the product of the constant term and coefficient of ${x^2}$ is a large number with a large number of divisors. Special care should be taken in such cases.