Let's factor the polynomial expression [tex]\(16 y^4 - 256 x^{12}\)[/tex].
Step 1: Look for a common factor in the terms of the expression. We can see that both terms share a factor of 16:
[tex]\[ 16 y^4 - 256 x^{12} = 16(y^4 - 16 x^{12}) \][/tex]
Step 2: Now we need to factor the polynomial inside the parentheses, [tex]\(y^4 - 16 x^{12}\)[/tex]. Notice that both [tex]\(y^4\)[/tex] and [tex]\(16 x^{12}\)[/tex] are perfect squares:
[tex]\[ y^4 = (y^2)^2 \][/tex]
[tex]\[ 16 x^{12} = (4 x^6)^2 \][/tex]
Step 3: Recognize that this is a difference of squares of the form [tex]\(a^2 - b^2\)[/tex], which factors as [tex]\((a - b)(a + b)\)[/tex]. In this case:
[tex]\[ a = y^2 \][/tex]
[tex]\[ b = 4 x^6 \][/tex]
Therefore:
[tex]\[ y^4 - 16 x^{12} = (y^2 - 4 x^6)(y^2 + 4 x^6) \][/tex]
Step 4: Now let's plug this back into our original expression:
[tex]\[ 16(y^4 - 16 x^{12}) = 16(y^2 - 4 x^6)(y^2 + 4 x^6) \][/tex]
Step 5: We can notice that [tex]\(y^2 - 4 x^6\)[/tex] is also a difference of squares:
[tex]\[ y^2 - 4 x^6 = (y - 2 x^3)(y + 2 x^3) \][/tex]
Substitute this back in:
[tex]\[ 16(y^2 - 4 x^6)(y^2 + 4 x^6) = 16(y - 2 x^3)(y + 2 x^3)(y^2 + 4 x^6) \][/tex]
Finally, the fully factored form is:
[tex]\[ 16 y^4 - 256 x^{12} = -16 (2 x^3 - y)(2 x^3 + y)(4 x^6 + y^2) \][/tex]
This matches our original result. Therefore, the factored form is:
[tex]\[ 16 y^4-256 x^{12}= -16(2 x^3 - y)(2 x^3 + y)(4 x^6 + y^2) \][/tex]
