Question

How do you factor $4{{x}^{2}}-26$?
(a) Factor by grouping
(b) Zero putting
(c) Guessing the factors
(d) Using a formula

Answer 1

Hint: To find the factor of the given equation $4{{x}^{2}}-26$, we will try to factorize it by using the formula of ${{a}^{2}}-{{b}^{2}}$ . We will start off by taking the term 2 common which is coming as a common term among both the terms. Then we go on with setting the terms as squares of a term and then by using ${{a}^{2}}-{{b}^{2}}$ as $(a+b)(a-b)$ .

Complete step by step solution:
According to the question, our given equation is, $4{{x}^{2}}-26$ and we are to factorize this equation.
So, to start with,
$4{{x}^{2}}-26$
Now, we can easily see we can take common 2 from both terms,
$=2(2{{x}^{2}}-13)$
One has to determine all the terms that were multiplied to obtain the given polynomial. Then try to factor every term that you got in the first step and this continues until you cannot factor further. When you can’t perform any more factoring, it is said that the polynomial is factored completely.
Again, we can factorize $2{{x}^{2}}-13$ as ${{(\sqrt{2}x)}^{2}}-{{(\sqrt{13})}^{2}}$ ,
So, now we do have, ${{a}^{2}}-{{b}^{2}}$ as $(a+b)(a-b)$ by using algebraic formulas.
Hence,
${{(\sqrt{2}x)}^{2}}-{{(\sqrt{13})}^{2}}$can be written as $(\sqrt{2}x+\sqrt{13})(\sqrt{2}x-\sqrt{13})$
Then factoring $4{{x}^{2}}-26$we are getting, $2(\sqrt{2}x+\sqrt{13})(\sqrt{2}x-\sqrt{13})$.
For any factorization we will always try to make the coefficient as an integer. In this problem, we are not finding any factors with an integer coefficient , that is why we are using an irrational coefficient in this problem.
Hence the solution is, (d) Using a formula.

Note: A polynomial $4{{x}^{2}}-26$ can be written as a product of two or more polynomials of degree less than or equal to that of it. Each polynomial involved in the product will be a factor of it. Monomials can be factorized in the same way as integers, just by writing the monomial as the product of its constituent prime factors. In the case of monomials, these prime factors can be integers as well as other monomials which cannot be factorized further.