Answer :
To factor the expression [tex]\(x^4 + 64y^4\)[/tex], we can use some algebraic techniques. Let's go through the steps to factor this polynomial:
Step 1: Recognize the form of the polynomial.
The given polynomial is [tex]\(x^4 + 64y^4\)[/tex], which resembles the sum of squares but in higher degree terms.
Step 2: Use algebraic identities for sum of squares in higher degree polynomials.
We know that [tex]\(x^4 + 64y^4\)[/tex] can be factored using the sum of powers formula for quartic terms. In this case, it can be factored into the product of two polynomials of the form [tex]\((a^2 - 2ab + b^2)(a^2 + 2ab + b^2)\)[/tex], where [tex]\(a = x^2\)[/tex] and [tex]\(b = 4y^2\)[/tex].
Step 3: Apply the general factorization:
We rewrite the polynomial as:
[tex]\[ x^4 + 64y^4 = (x^2)^2 + (4y^2)^2 \][/tex]
This can be factored using a specific identity for sum of squares for quartic terms.
[tex]\[ x^4 + 64y^4 = (x^2 - 4xy + 8y^2)(x^2 + 4xy + 8y^2) \][/tex]
Step 4: Verify the factorization.
Let’s expand the factored form to verify it matches the original expression:
[tex]\[ (x^2 - 4xy + 8y^2)(x^2 + 4xy + 8y^2) \][/tex]
Expanding this step by step:
[tex]\[ = x^2(x^2 + 4xy + 8y^2) - 4xy(x^2 + 4xy + 8y^2) + 8y^2(x^2 + 4xy + 8y^2) \][/tex]
This expands to:
[tex]\[ = x^4 + 4x^3y + 8x^2y^2 - 4x^3y - 16x^2y^2 - 32xy^3 + 8x^2y^2 + 32xy^3 + 64y^4 \][/tex]
Combining like terms:
[tex]\[ = x^4 + (4x^3y - 4x^3y) + (8x^2y^2 - 16x^2y^2 + 8x^2y^2) + (32xy^3 - 32xy^3) + 64y^4 = x^4 + 64y^4 \][/tex]
Thus, our factorization is correct, and we conclude that:
[tex]\[ x^4 + 64y^4 = (x^2 - 4xy + 8y^2)(x^2 + 4xy + 8y^2) \][/tex]
This is the factorized form of the given polynomial.
Step 1: Recognize the form of the polynomial.
The given polynomial is [tex]\(x^4 + 64y^4\)[/tex], which resembles the sum of squares but in higher degree terms.
Step 2: Use algebraic identities for sum of squares in higher degree polynomials.
We know that [tex]\(x^4 + 64y^4\)[/tex] can be factored using the sum of powers formula for quartic terms. In this case, it can be factored into the product of two polynomials of the form [tex]\((a^2 - 2ab + b^2)(a^2 + 2ab + b^2)\)[/tex], where [tex]\(a = x^2\)[/tex] and [tex]\(b = 4y^2\)[/tex].
Step 3: Apply the general factorization:
We rewrite the polynomial as:
[tex]\[ x^4 + 64y^4 = (x^2)^2 + (4y^2)^2 \][/tex]
This can be factored using a specific identity for sum of squares for quartic terms.
[tex]\[ x^4 + 64y^4 = (x^2 - 4xy + 8y^2)(x^2 + 4xy + 8y^2) \][/tex]
Step 4: Verify the factorization.
Let’s expand the factored form to verify it matches the original expression:
[tex]\[ (x^2 - 4xy + 8y^2)(x^2 + 4xy + 8y^2) \][/tex]
Expanding this step by step:
[tex]\[ = x^2(x^2 + 4xy + 8y^2) - 4xy(x^2 + 4xy + 8y^2) + 8y^2(x^2 + 4xy + 8y^2) \][/tex]
This expands to:
[tex]\[ = x^4 + 4x^3y + 8x^2y^2 - 4x^3y - 16x^2y^2 - 32xy^3 + 8x^2y^2 + 32xy^3 + 64y^4 \][/tex]
Combining like terms:
[tex]\[ = x^4 + (4x^3y - 4x^3y) + (8x^2y^2 - 16x^2y^2 + 8x^2y^2) + (32xy^3 - 32xy^3) + 64y^4 = x^4 + 64y^4 \][/tex]
Thus, our factorization is correct, and we conclude that:
[tex]\[ x^4 + 64y^4 = (x^2 - 4xy + 8y^2)(x^2 + 4xy + 8y^2) \][/tex]
This is the factorized form of the given polynomial.