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Answer :
Sure! Let's factor the polynomial expression [tex]\( 16y^4 - 625x^4 \)[/tex] step by step.
1. Recognize that this is a difference of squares:
The expression [tex]\( 16y^4 - 625x^4 \)[/tex] is a difference of squares because it can be written in the form [tex]\( a^2 - b^2 \)[/tex].
Here, [tex]\( a^2 = 16y^4 \)[/tex] and [tex]\( b^2 = 625x^4 \)[/tex].
2. Find the square roots of the terms:
To factor the expression as a difference of squares, we need the square roots of [tex]\( a^2 \)[/tex] and [tex]\( b^2 \)[/tex].
- The square root of [tex]\( 16y^4 \)[/tex] is [tex]\( 4y^2 \)[/tex].
- The square root of [tex]\( 625x^4 \)[/tex] is [tex]\( 25x^2 \)[/tex].
So, our expression can be written as:
[tex]\( (4y^2)^2 - (25x^2)^2 \)[/tex].
3. Apply the difference of squares formula:
Recall that the difference of squares formula is [tex]\( a^2 - b^2 = (a - b)(a + b) \)[/tex].
Using the square roots we found:
[tex]\[
16y^4 - 625x^4 = (4y^2 - 25x^2)(4y^2 + 25x^2)
\][/tex]
4. Factor further if possible:
Notice that [tex]\( 4y^2 - 25x^2 \)[/tex] is itself a difference of squares:
[tex]\[
4y^2 - 25x^2 = (2y)^2 - (5x)^2 = (2y - 5x)(2y + 5x)
\][/tex]
Hence, the complete factorization of the original expression is:
[tex]\[
(4y^2 - 25x^2)(4y^2 + 25x^2) = (2y - 5x)(2y + 5x)(4y^2 + 25x^2)
\][/tex]
Therefore, the fully factored form of the polynomial expression [tex]\( 16y^4 - 625x^4 \)[/tex] is:
[tex]\[
(2y - 5x)(2y + 5x)(4y^2 + 25x^2)
\][/tex]
1. Recognize that this is a difference of squares:
The expression [tex]\( 16y^4 - 625x^4 \)[/tex] is a difference of squares because it can be written in the form [tex]\( a^2 - b^2 \)[/tex].
Here, [tex]\( a^2 = 16y^4 \)[/tex] and [tex]\( b^2 = 625x^4 \)[/tex].
2. Find the square roots of the terms:
To factor the expression as a difference of squares, we need the square roots of [tex]\( a^2 \)[/tex] and [tex]\( b^2 \)[/tex].
- The square root of [tex]\( 16y^4 \)[/tex] is [tex]\( 4y^2 \)[/tex].
- The square root of [tex]\( 625x^4 \)[/tex] is [tex]\( 25x^2 \)[/tex].
So, our expression can be written as:
[tex]\( (4y^2)^2 - (25x^2)^2 \)[/tex].
3. Apply the difference of squares formula:
Recall that the difference of squares formula is [tex]\( a^2 - b^2 = (a - b)(a + b) \)[/tex].
Using the square roots we found:
[tex]\[
16y^4 - 625x^4 = (4y^2 - 25x^2)(4y^2 + 25x^2)
\][/tex]
4. Factor further if possible:
Notice that [tex]\( 4y^2 - 25x^2 \)[/tex] is itself a difference of squares:
[tex]\[
4y^2 - 25x^2 = (2y)^2 - (5x)^2 = (2y - 5x)(2y + 5x)
\][/tex]
Hence, the complete factorization of the original expression is:
[tex]\[
(4y^2 - 25x^2)(4y^2 + 25x^2) = (2y - 5x)(2y + 5x)(4y^2 + 25x^2)
\][/tex]
Therefore, the fully factored form of the polynomial expression [tex]\( 16y^4 - 625x^4 \)[/tex] is:
[tex]\[
(2y - 5x)(2y + 5x)(4y^2 + 25x^2)
\][/tex]
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