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

**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 \[ 8x^2y^2+4bxy^2+\frac{1}{2}b^2y^2. \]