To solve the expression [tex]\(\left(3 x^{4a-5} + 2 y^{2a+4}\right)^2\)[/tex], we'll follow a detailed, step-by-step expansion using algebraic principles. Specifically, we will apply the square of a binomial formula:
[tex]\[
(A + B)^2 = A^2 + 2AB + B^2
\][/tex]
Let's identify our [tex]\(A\)[/tex] and [tex]\(B\)[/tex] in the expression:
- [tex]\(A = 3 x^{4a-5}\)[/tex]
- [tex]\(B = 2 y^{2a+4}\)[/tex]
Now, let's apply the formula step-by-step.
1. Square the first term (A):
[tex]\[
A^2 = (3 x^{4a-5})^2 = 9 x^{2(4a-5)} = 9 x^{8a-10}
\][/tex]
2. Square the second term (B):
[tex]\[
B^2 = (2 y^{2a+4})^2 = 4 y^{2(2a+4)} = 4 y^{4a+8}
\][/tex]
3. Multiply the two terms together and double the product:
[tex]\[
2AB = 2 \cdot (3 x^{4a-5}) \cdot (2 y^{2a+4}) = 2 \cdot 3 \cdot 2 \cdot x^{4a-5} \cdot y^{2a+4} = 12 x^{4a-5} y^{2a+4}
\][/tex]
Now, we combine all these expanded parts:
[tex]\[
\left(3 x^{4a-5} + 2 y^{2a+4}\right)^2 = A^2 + 2AB + B^2 = 9 x^{8a-10} + 12 x^{4a-5} y^{2a+4} + 4 y^{4a+8}
\][/tex]
Thus, the expanded form of the given expression is:
[tex]\[
12 x^{4a-5} y^{2a+4} + 9 x^{8a-10} + 4 y^{4a+8}
\][/tex]
