Answer :
Certainly! Let's factor the expression [tex]\(4x^2 - 81y^4\)[/tex] step-by-step.
1. Recognize the form of the expression:
The expression [tex]\(4x^2 - 81y^4\)[/tex] can be noticed as a difference of squares because both terms are perfect squares. Here's how we can see that:
[tex]\(4x^2\)[/tex] is [tex]\( (2x)^2 \)[/tex]
[tex]\(81y^4\)[/tex] is [tex]\( (9y^2)^2 \)[/tex]
So, the expression [tex]\(4x^2 - 81y^4\)[/tex] can be rewritten as:
[tex]\[ (2x)^2 - (9y^2)^2 \][/tex]
2. Apply the difference of squares formula:
The difference of squares formula states that:
[tex]\[ a^2 - b^2 = (a - b)(a + b) \][/tex]
Here, let [tex]\(a = 2x\)[/tex] and [tex]\(b = 9y^2\)[/tex].
3. Substitute [tex]\(a\)[/tex] and [tex]\(b\)[/tex] into the formula:
[tex]\[ (2x)^2 - (9y^2)^2 = (2x - 9y^2)(2x + 9y^2) \][/tex]
Therefore, the factored form of [tex]\(4x^2 - 81y^4\)[/tex] is:
[tex]\[ (2x - 9y^2)(2x + 9y^2) \][/tex]
This completes the factorization of the given expression.
1. Recognize the form of the expression:
The expression [tex]\(4x^2 - 81y^4\)[/tex] can be noticed as a difference of squares because both terms are perfect squares. Here's how we can see that:
[tex]\(4x^2\)[/tex] is [tex]\( (2x)^2 \)[/tex]
[tex]\(81y^4\)[/tex] is [tex]\( (9y^2)^2 \)[/tex]
So, the expression [tex]\(4x^2 - 81y^4\)[/tex] can be rewritten as:
[tex]\[ (2x)^2 - (9y^2)^2 \][/tex]
2. Apply the difference of squares formula:
The difference of squares formula states that:
[tex]\[ a^2 - b^2 = (a - b)(a + b) \][/tex]
Here, let [tex]\(a = 2x\)[/tex] and [tex]\(b = 9y^2\)[/tex].
3. Substitute [tex]\(a\)[/tex] and [tex]\(b\)[/tex] into the formula:
[tex]\[ (2x)^2 - (9y^2)^2 = (2x - 9y^2)(2x + 9y^2) \][/tex]
Therefore, the factored form of [tex]\(4x^2 - 81y^4\)[/tex] is:
[tex]\[ (2x - 9y^2)(2x + 9y^2) \][/tex]
This completes the factorization of the given expression.