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Answer :
To factor the polynomial expression [tex]\(16y^4 - 625x^4\)[/tex], we can view it as a difference of squares. The difference of squares formula is [tex]\(a^2 - b^2 = (a - b)(a + b)\)[/tex].
Let's start by rewriting the expression in terms of squares:
1. Notice that [tex]\(16y^4\)[/tex] is a perfect square:
[tex]\[
16y^4 = (2y^2)^2
\][/tex]
2. Similarly, [tex]\(625x^4\)[/tex] is a perfect square:
[tex]\[
625x^4 = (5x^2)^2
\][/tex]
3. Now apply the difference of squares formula:
[tex]\[
16y^4 - 625x^4 = (2y^2)^2 - (5x^2)^2 = (2y^2 - 5x^2)(2y^2 + 5x^2)
\][/tex]
However, we notice that the expression [tex]\(2y^2 - 5x^2\)[/tex] can also be factored further using the difference of squares formula again:
4. Realize that [tex]\(4y^4 - 25x^4\)[/tex] can be split as:
[tex]\[
(2y - 5x)(2y + 5x)
\][/tex]
Putting everything together, we have:
- The expression [tex]\(16y^4 - 625x^4\)[/tex] factors into:
[tex]\[
-(5x - 2y)(5x + 2y)(25x^2 + 4y^2)
\][/tex]
Therefore, the factorization of the polynomial expression [tex]\(16y^4 - 625x^4\)[/tex] is:
[tex]\[
-(5x - 2y)(5x + 2y)(25x^2 + 4y^2)
\][/tex]
Let's start by rewriting the expression in terms of squares:
1. Notice that [tex]\(16y^4\)[/tex] is a perfect square:
[tex]\[
16y^4 = (2y^2)^2
\][/tex]
2. Similarly, [tex]\(625x^4\)[/tex] is a perfect square:
[tex]\[
625x^4 = (5x^2)^2
\][/tex]
3. Now apply the difference of squares formula:
[tex]\[
16y^4 - 625x^4 = (2y^2)^2 - (5x^2)^2 = (2y^2 - 5x^2)(2y^2 + 5x^2)
\][/tex]
However, we notice that the expression [tex]\(2y^2 - 5x^2\)[/tex] can also be factored further using the difference of squares formula again:
4. Realize that [tex]\(4y^4 - 25x^4\)[/tex] can be split as:
[tex]\[
(2y - 5x)(2y + 5x)
\][/tex]
Putting everything together, we have:
- The expression [tex]\(16y^4 - 625x^4\)[/tex] factors into:
[tex]\[
-(5x - 2y)(5x + 2y)(25x^2 + 4y^2)
\][/tex]
Therefore, the factorization of the polynomial expression [tex]\(16y^4 - 625x^4\)[/tex] is:
[tex]\[
-(5x - 2y)(5x + 2y)(25x^2 + 4y^2)
\][/tex]
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