Factor completely [tex]\( 49x^2 - 81 \)[/tex].

A. [tex]\( (7x + 9)^2 \)[/tex]
B. [tex]\( (7x - 9)^2 \)[/tex]
C. [tex]\( (7x + 9)(7x - 9) \)[/tex]
D. [tex]\( 7x - 9 \)[/tex]



Answer :

To factor the expression [tex]\(49 x^2 - 81\)[/tex] completely, we can recognize that this is a difference of squares. The difference of squares is a common algebraic pattern that can be factored using the formula:

[tex]\[ a^2 - b^2 = (a - b)(a + b) \][/tex]

In this case, we need to identify [tex]\(a\)[/tex] and [tex]\(b\)[/tex] such that [tex]\(49 x^2\)[/tex] and [tex]\(81\)[/tex] are perfect squares.

1. Notice that [tex]\(49 x^2\)[/tex] is a perfect square:
[tex]\[ 49 x^2 = (7x)^2 \][/tex]
So [tex]\(a = 7x\)[/tex].

2. Notice that [tex]\(81\)[/tex] is also a perfect square:
[tex]\[ 81 = 9^2 \][/tex]
So [tex]\(b = 9\)[/tex].

Now we apply the difference of squares formula:

[tex]\[ 49 x^2 - 81 = (7x)^2 - 9^2 = (7x - 9)(7x + 9) \][/tex]

Thus, the factorization of [tex]\(49 x^2 - 81\)[/tex] is:

[tex]\[ (7x - 9)(7x + 9) \][/tex]

So, the completely factored form of [tex]\(49 x^2 - 81\)[/tex] is:
[tex]\[ (7x - 9)(7x + 9) \][/tex]