2.1.2) $$ (2 a+3)\left(4 a^{2}-6 a+9\right) $$ 2.13) $$ \frac{1}{2}(4 x y+b y)^{2} $$

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**Problem 2.1.2**

We start with expanding
$$
(2a+3)(4a^2-6a+9).
$$

1. Distribute each term of the first factor:
   - Multiply $$2a$$ by each term of the second factor:
     $$
     \begin{aligned}
     2a \times 4a^2 &= 8a^3,\[1mm]
     2a \times (-6a) &= -12a^2,\[1mm]
     2a \times 9 &= 18a.
     \end{aligned}
     $$
   - Multiply $$3$$ by each term of the second factor:
     $$
     \begin{aligned}
     3 \times 4a^2 &= 12a^2,\[1mm]
     3 \times (-6a) &= -18a,\[1mm]
     3 \times 9 &= 27.
     \end{aligned}
     $$

2. Combine like terms:
   $$
   \begin{aligned}
   8a^3 &: \text{(only term)}\[1mm]
   -12a^2+12a^2 &= 0,\[1mm]
   18a-18a &= 0,\[1mm]
   27 &: \text{(constant term)}
   \end{aligned}
   $$

Thus, the expanded form is
$$
8a^3+27.
$$

---

**Problem 2.13**

We need to simplify
$$
\frac{1}{2}(4xy+by)^2.
$$

1. First, expand the squared binomial using the formula $$ (p+q)^2 = p^2+2pq+q^2 $$ with $$ p=4xy $$ and $$ q=by $$:
   $$
   \begin{aligned}
   (4xy+by)^2 &= (4xy)^2+2\cdot(4xy)(by)+(by)^2\[1mm]
   &= 16x^2y^2+8bxy^2+b^2y^2.
   \end{aligned}
   $$

2. Now, multiply the entire expression by $$\frac{1}{2}$$:
   $$
   \begin{aligned}
   \frac{1}{2}(16x^2y^2) &= 8x^2y^2,\[1mm]
   \frac{1}{2}(8bxy^2) &= 4bxy^2,\[1mm]
   \frac{1}{2}(b^2y^2) &= \frac{1}{2}b^2y^2.
   \end{aligned}
   $$

Thus, the simplified form is
$$
8x^2y^2+4bxy^2+\frac{1}{2}b^2y^2.
$$
**Problem 2.1.2** $$ (2a+3)(4a^2-6a+9) = 8a^3 + 27 $$ **Problem 2.13** $$ \frac{1}{2}(4xy + by)^2 = 8x^2y^2 + 4bxy^2 + \frac{1}{2}b^2y^2 $$

2.1.2) \( (2 a+3)\left(4 a^{2}-6 a+9\right) \) 2.13) \( \frac{1}{2}(4 x y+b y)^{2} \)

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