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Answer :
Let's factor the polynomial expression [tex]\(16y^4 - 625x^4\)[/tex].
The expression is in the form of a difference of squares. The difference of squares formula is:
[tex]\[
a^2 - b^2 = (a - b)(a + b)
\][/tex]
First, we need to identify [tex]\(a\)[/tex] and [tex]\(b\)[/tex] such that:
1. [tex]\(a^2 = 16y^4\)[/tex]
2. [tex]\(b^2 = 625x^4\)[/tex]
We find [tex]\(a\)[/tex] and [tex]\(b\)[/tex] by taking the square roots:
- For [tex]\(a\)[/tex], we recognize [tex]\(16y^4\)[/tex] as [tex]\((4y^2)^2\)[/tex]. Thus, [tex]\(a = 4y^2\)[/tex].
- For [tex]\(b\)[/tex], we recognize [tex]\(625x^4\)[/tex] as [tex]\((25x^2)^2\)[/tex]. Thus, [tex]\(b = 25x^2\)[/tex].
Now, apply the difference of squares formula:
[tex]\[
16y^4 - 625x^4 = (4y^2)^2 - (25x^2)^2 = (4y^2 - 25x^2)(4y^2 + 25x^2)
\][/tex]
Next, let's check if the first expression, [tex]\(4y^2 - 25x^2\)[/tex], can be further factored. Notice that:
- [tex]\(4y^2\)[/tex] is [tex]\((2y)^2\)[/tex] and
- [tex]\(25x^2\)[/tex] is [tex]\((5x)^2\)[/tex],
So, [tex]\(4y^2 - 25x^2\)[/tex] is also a difference of squares:
[tex]\[
4y^2 - 25x^2 = (2y - 5x)(2y + 5x)
\][/tex]
The second expression, [tex]\(4y^2 + 25x^2\)[/tex], does not factor further using real numbers because it's a sum and not a difference.
Putting everything together, the factorization of [tex]\(16y^4 - 625x^4\)[/tex] is:
[tex]\[
(2y - 5x)(2y + 5x)(4y^2 + 25x^2)
\][/tex]
This is the fully factored form of the given polynomial.
The expression is in the form of a difference of squares. The difference of squares formula is:
[tex]\[
a^2 - b^2 = (a - b)(a + b)
\][/tex]
First, we need to identify [tex]\(a\)[/tex] and [tex]\(b\)[/tex] such that:
1. [tex]\(a^2 = 16y^4\)[/tex]
2. [tex]\(b^2 = 625x^4\)[/tex]
We find [tex]\(a\)[/tex] and [tex]\(b\)[/tex] by taking the square roots:
- For [tex]\(a\)[/tex], we recognize [tex]\(16y^4\)[/tex] as [tex]\((4y^2)^2\)[/tex]. Thus, [tex]\(a = 4y^2\)[/tex].
- For [tex]\(b\)[/tex], we recognize [tex]\(625x^4\)[/tex] as [tex]\((25x^2)^2\)[/tex]. Thus, [tex]\(b = 25x^2\)[/tex].
Now, apply the difference of squares formula:
[tex]\[
16y^4 - 625x^4 = (4y^2)^2 - (25x^2)^2 = (4y^2 - 25x^2)(4y^2 + 25x^2)
\][/tex]
Next, let's check if the first expression, [tex]\(4y^2 - 25x^2\)[/tex], can be further factored. Notice that:
- [tex]\(4y^2\)[/tex] is [tex]\((2y)^2\)[/tex] and
- [tex]\(25x^2\)[/tex] is [tex]\((5x)^2\)[/tex],
So, [tex]\(4y^2 - 25x^2\)[/tex] is also a difference of squares:
[tex]\[
4y^2 - 25x^2 = (2y - 5x)(2y + 5x)
\][/tex]
The second expression, [tex]\(4y^2 + 25x^2\)[/tex], does not factor further using real numbers because it's a sum and not a difference.
Putting everything together, the factorization of [tex]\(16y^4 - 625x^4\)[/tex] is:
[tex]\[
(2y - 5x)(2y + 5x)(4y^2 + 25x^2)
\][/tex]
This is the fully factored form of the given polynomial.
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