Given:
[tex]1-49a^2[/tex]
Required:
We need to factorize the given expression.
Explanation:
[tex]Use\text{ 1=1}^2\text{ and }49=7^2.[/tex]
[tex]1-49a^2=1^2-7^2a^2[/tex][tex]Use\text{ }7^2a^2=(7a)^2.[/tex]
[tex]1-49a^2=1^2-(7a)^2[/tex]
The given expression is a difference of perfect sqaure because 1 is the perfect sqaure and 49 is also the perfect square of 7.
We know that a multiple of a perfect sqaure is also a perfect square.
[tex]49a^2\text{ is a perfect square.}[/tex]
We get
[tex]1^2-(7a)^2,[/tex]
Use the structure of the difference of perfect squares to factorize the given expression.
[tex](a^2-b^2)=(a-b)(a+b)[/tex]
Substitute a =1 and b =7a in the formula.
[tex]1^2-(7a)^2=(1-7a)(1+7a)[/tex]
[tex]1-49a^2=(1-7a)(1+7a)[/tex]
The number of factors = 2
Final answer:
The given expression is a difference of perfect sqaure because 1 is the perfect sqaure and 49 is also the perfect square of 7.
We know that a multiple of a perfect sqaure is also a perfect square.
[tex]1-49a^2=(1-7a)(1+7a)[/tex]
The number of factors = 2
