1. **Problem statement:** Write the given expressions as the square of a binomial. 2. **Formula used:** A perfect square trinomial can be written as $$(ax + b)^2 = a^2x^2 + 2abx + b^2$$ 3. **Step-by-step solutions:** **a)** $4x^2 + 12x + 9$ - Recognize $4x^2 = (2x)^2$, $9 = 3^2$, and middle term $12x = 2 \cdot 2x \cdot 3$ - So, $4x^2 + 12x + 9 = (2x + 3)^2$ **b)** $x^2 - 14x + 49$ - Recognize $x^2 = (x)^2$, $49 = 7^2$, and middle term $-14x = 2 \cdot x \cdot (-7)$ - So, $x^2 - 14x + 49 = (x - 7)^2$ **c)** $16x^2 + 16x + 4$ - Recognize $16x^2 = (4x)^2$, $4 = 2^2$, and middle term $16x = 2 \cdot 4x \cdot 2$ - So, $16x^2 + 16x + 4 = (4x + 2)^2$ **d)** $9 - 24x + 16x^2$ - Rewrite as $16x^2 - 24x + 9$ - Recognize $16x^2 = (4x)^2$, $9 = 3^2$, and middle term $-24x = 2 \cdot 4x \cdot (-3)$ - So, $9 - 24x + 16x^2 = (4x - 3)^2$ **e)** $16x^2 - 40xy + 25y^2$ - Recognize $16x^2 = (4x)^2$, $25y^2 = (5y)^2$, and middle term $-40xy = 2 \cdot 4x \cdot (-5y)$ - So, $16x^2 - 40xy + 25y^2 = (4x - 5y)^2$ **f)** $4x^2 + 4xy + y^2$ - Recognize $4x^2 = (2x)^2$, $y^2 = (y)^2$, and middle term $4xy = 2 \cdot 2x \cdot y$ - So, $4x^2 + 4xy + y^2 = (2x + y)^2$ **g)** $36x^2 + 24xy + 4y^2$ - Recognize $36x^2 = (6x)^2$, $4y^2 = (2y)^2$, and middle term $24xy = 2 \cdot 6x \cdot 2y$ - So, $36x^2 + 24xy + 4y^2 = (6x + 2y)^2$ **h)** $81x^2 + 90xy + 25y^2$ - Recognize $81x^2 = (9x)^2$, $25y^2 = (5y)^2$, and middle term $90xy = 2 \cdot 9x \cdot 5y$ - So, $81x^2 + 90xy + 25y^2 = (9x + 5y)^2$ **i)** $\frac{x^2}{4} + xy + y^2$ - Rewrite as $(\frac{x}{2})^2 + xy + y^2$ - Check middle term: $xy = 2 \cdot \frac{x}{2} \cdot y$ - So, $\frac{x^2}{4} + xy + y^2 = \left(\frac{x}{2} + y\right)^2$ **j)** $\frac{x^2}{9} + \frac{xy}{6} + \frac{y^2}{16}$ - Rewrite as $(\frac{x}{3})^2 + \frac{xy}{6} + (\frac{y}{4})^2$ - Check middle term: $\frac{xy}{6} = 2 \cdot \frac{x}{3} \cdot \frac{y}{4}$ since $2 \cdot \frac{1}{3} \cdot \frac{1}{4} = \frac{2}{12} = \frac{1}{6}$ - So, $\frac{x^2}{9} + \frac{xy}{6} + \frac{y^2}{16} = \left(\frac{x}{3} + \frac{y}{4}\right)^2$ 4. **Final answers:** - a) $(2x + 3)^2$ - b) $(x - 7)^2$ - c) $(4x + 2)^2$ - d) $(4x - 3)^2$ - e) $(4x - 5y)^2$ - f) $(2x + y)^2$ - g) $(6x + 2y)^2$ - h) $(9x + 5y)^2$ - i) $\left(\frac{x}{2} + y\right)^2$ - j) $\left(\frac{x}{3} + \frac{y}{4}\right)^2$