# Prove that the factor of ${(2n + 3)^2} - {(2n - 3)^2}$ is 8, where n is a natural number.

Last updated date: 20th Mar 2023

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Answer

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Hint: Use algebra to solve the expression by applying the general formula of $a^2 - b^2$.

We have given the expression ${(2n + 3)^2} - {(2n - 3)^2}$ . Observe its in the form of ${a^2} - {b^2}$ where, $a = {(2n + 3)^2},b = {(2n - 3)^2}$ and we can use the formula ${a^2} - {b^2} = (a + b)(a - b)$ . Using the formula, we’ll get

$\

{(2n + 3)^2} - {(2n - 3)^2} \\

= (2n + 3 + 2n - 3)(2n + 3 - 2n + 3){\text{ [}}{a^2} - {b^2} = (a + b)(a - b){\text{]}} \\

{\text{ = (4n)(6)}} \\

{\text{ = 24n}} \\

\ $

Since, it’s nothing but 24n so we can easily write it in terms of 8 as $24n = 8(3n)$ and hence we can conclude that 8 is the factor of given expression.

Note: In this question, just follow it and using the fundamentals of algebra, we can solve it but we need to be careful with the calculation part.

We have given the expression ${(2n + 3)^2} - {(2n - 3)^2}$ . Observe its in the form of ${a^2} - {b^2}$ where, $a = {(2n + 3)^2},b = {(2n - 3)^2}$ and we can use the formula ${a^2} - {b^2} = (a + b)(a - b)$ . Using the formula, we’ll get

$\

{(2n + 3)^2} - {(2n - 3)^2} \\

= (2n + 3 + 2n - 3)(2n + 3 - 2n + 3){\text{ [}}{a^2} - {b^2} = (a + b)(a - b){\text{]}} \\

{\text{ = (4n)(6)}} \\

{\text{ = 24n}} \\

\ $

Since, it’s nothing but 24n so we can easily write it in terms of 8 as $24n = 8(3n)$ and hence we can conclude that 8 is the factor of given expression.

Note: In this question, just follow it and using the fundamentals of algebra, we can solve it but we need to be careful with the calculation part.

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