To expand $(2x - 3)^2$ using the binomial theorem:

The binomial theorem states:

$(a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$

Here:

• $a = 2x$
• $b = -3$
• $n = 2$

The expansion for $(2x - 3)^2$ is:

$(2x - 3)^2 = \binom{2}{0}(2x)^2(-3)^0 + \binom{2}{1}(2x)^1(-3)^1 + \binom{2}{2}(2x)^0(-3)^2$

Step-by-step calculation:

1. $\binom{2}{0}(2x)^2(-3)^0 = 1 \cdot (2x)^2 \cdot 1 = 4x^2$
2. $\binom{2}{1}(2x)^1(-3)^1 = 2 \cdot (2x) \cdot (-3) = -12x$
3. $\binom{2}{2}(2x)^0(-3)^2 = 1 \cdot 1 \cdot 9 = 9$

Final result:

$(2x - 3)^2 = 4x^2 - 12x + 9$

Let me know if you'd like further explanation!

Related Questions:

1. How do you generalize the binomial theorem for higher powers?
2. What are the properties of binomial coefficients $\binom{n}{k}$?
3. How can the binomial theorem be applied to negative or fractional exponents?
4. What is the geometric interpretation of $(2x - 3)^2$?
5. How does this method differ from direct multiplication of $(2x - 3)(2x - 3)$?

Tip:

Always verify binomial expansions by multiplying the terms directly as a cross-check!