Sure, let's solve each of these expressions step-by-step.
### a) [tex]\((9 + 4m)^2\)[/tex]
To expand [tex]\((9 + 4m)^2\)[/tex], we use the binomial expansion formula [tex]\((a + b)^2 = a^2 + 2ab + b^2\)[/tex], where [tex]\(a = 9\)[/tex] and [tex]\(b = 4m\)[/tex].
[tex]\[ a^2 = 9^2 = 81 \][/tex]
[tex]\[ 2ab = 2 \cdot 9 \cdot 4m = 72m \][/tex]
[tex]\[ b^2 = (4m)^2 = 16m^2 \][/tex]
Thus, [tex]\((9 + 4m)^2\)[/tex] expands to:
[tex]\[ 16m^2 + 72m + 81 \][/tex]
### b) [tex]\((2x - 3z)^2\)[/tex]
To expand [tex]\((2x - 3z)^2\)[/tex], we use the binomial expansion formula [tex]\((a - b)^2 = a^2 - 2ab + b^2\)[/tex], where [tex]\(a = 2x\)[/tex] and [tex]\(b = 3z\)[/tex].
[tex]\[ a^2 = (2x)^2 = 4x^2 \][/tex]
[tex]\[ 2ab = 2 \cdot 2x \cdot 3z = 12xz \][/tex]
[tex]\[ b^2 = (3z)^2 = 9z^2 \][/tex]
Thus, [tex]\((2x - 3z)^2\)[/tex] expands to:
[tex]\[ 4x^2 - 12xz + 9z^2 \][/tex]
### c) [tex]\(\left(\frac{3}{6}w - \frac{y}{2}\right)^2\)[/tex]
First, let's simplify the terms inside the parentheses:
[tex]\[ \frac{3}{6}w = \frac{1}{2}w \][/tex]
So, the expression becomes:
[tex]\[ \left(\frac{1}{2}w - \frac{y}{2}\right)^2 \][/tex]
To expand this, we use the binomial expansion formula [tex]\((a - b)^2 = a^2 - 2ab + b^2\)[/tex], where [tex]\(a = \frac{1}{2}w\)[/tex] and [tex]\(b = \frac{1}{2}y\)[/tex].
[tex]\[ a^2 = \left(\frac{1}{2}w\right)^2 = \frac{1}{4}w^2 \][/tex]
[tex]\[ 2ab = 2 \cdot \frac{1}{2}w \cdot \frac{1}{2}y = \frac{1}{2}wy \][/tex]
[tex]\[ b^2 = \left(\frac{1}{2}y\right)^2 = \frac{1}{4}y^2 \][/tex]
Thus, [tex]\(\left(\frac{1}{2}w - \frac{y}{2}\right)^2\)[/tex] expands to:
[tex]\[ \frac{1}{4}w^2 - \frac{1}{2}wy + \frac{1}{4}y^2 \][/tex]
Therefore, the expanded forms are:
[tex]\[
\text{a) } (9 + 4m)^2 = 16m^2 + 72m + 81 \\
\text{b) } (2x - 3z)^2 = 4x^2 - 12xz + 9z^2 \\
\text{c) } \left(\frac{3}{6}w - \frac{y}{2}\right)^2 = \frac{1}{4}w^2 - \frac{1}{2}wy + \frac{1}{4}y^2
\][/tex]
