Answer :
Sure! Let's work through the detailed steps of multiplying the binomials [tex]\((7x^2 - 3y^2)(x^2 - 8y^2)\)[/tex] and simplifying the answer.
1. Distribute each term in the first binomial to each term in the second binomial:
[tex]\[ (7x^2 - 3y^2)(x^2 - 8y^2) = 7x^2 \cdot x^2 + 7x^2 \cdot (-8y^2) + (-3y^2) \cdot x^2 + (-3y^2) \cdot (-8y^2) \][/tex]
2. Calculate each product:
- [tex]\(7x^2 \cdot x^2 = 7x^4\)[/tex]
- [tex]\(7x^2 \cdot (-8y^2) = -56x^2y^2\)[/tex]
- [tex]\((-3y^2) \cdot x^2 = -3x^2y^2\)[/tex]
- [tex]\((-3y^2) \cdot (-8y^2) = 24y^4\)[/tex]
3. Combine like terms (terms with the same variables and exponents):
Combining [tex]\(-56x^2y^2\)[/tex] and [tex]\(-3x^2y^2\)[/tex]:
[tex]\[ -56x^2y^2 + (-3x^2y^2) = -59x^2y^2 \][/tex]
4. Write out the simplified expression by combining all of the products:
[tex]\[ 7x^4 - 59x^2y^2 + 24y^4 \][/tex]
Therefore, the expanded and simplified form of the product of the binomials [tex]\((7x^2 - 3y^2)(x^2 - 8y^2)\)[/tex] is:
[tex]\[ 7x^4 - 59x^2y^2 + 24y^4 \][/tex]
1. Distribute each term in the first binomial to each term in the second binomial:
[tex]\[ (7x^2 - 3y^2)(x^2 - 8y^2) = 7x^2 \cdot x^2 + 7x^2 \cdot (-8y^2) + (-3y^2) \cdot x^2 + (-3y^2) \cdot (-8y^2) \][/tex]
2. Calculate each product:
- [tex]\(7x^2 \cdot x^2 = 7x^4\)[/tex]
- [tex]\(7x^2 \cdot (-8y^2) = -56x^2y^2\)[/tex]
- [tex]\((-3y^2) \cdot x^2 = -3x^2y^2\)[/tex]
- [tex]\((-3y^2) \cdot (-8y^2) = 24y^4\)[/tex]
3. Combine like terms (terms with the same variables and exponents):
Combining [tex]\(-56x^2y^2\)[/tex] and [tex]\(-3x^2y^2\)[/tex]:
[tex]\[ -56x^2y^2 + (-3x^2y^2) = -59x^2y^2 \][/tex]
4. Write out the simplified expression by combining all of the products:
[tex]\[ 7x^4 - 59x^2y^2 + 24y^4 \][/tex]
Therefore, the expanded and simplified form of the product of the binomials [tex]\((7x^2 - 3y^2)(x^2 - 8y^2)\)[/tex] is:
[tex]\[ 7x^4 - 59x^2y^2 + 24y^4 \][/tex]