r. \( 22 y^{2}-9 y-1 \) v \( 9 a^{2}+30 a b+25 b^{2} \)

Sure, let's factor both expressions you've provided. --- ### **r. \( 22 y^{2} - 9 y - 1 \)** To factor the quadratic expression \( 22 y^{2} - 9 y - 1 \), we'll look for two numbers that multiply to \( 22 \times (-1) = -22 \) and add up to \( -9 \). 1. **Find the numbers:** - The numbers are **-11** and **2**, since: - \( -11 \times 2 = -22 \) - \( -11 + 2 = -9 \) 2. **Rewrite the middle term using these numbers:** \[ 22 y^{2} - 11 y + 2 y - 1 \] 3. **Factor by grouping:** \[ (22 y^{2} - 11 y) + (2 y - 1) = 11 y (2 y - 1) + 1 (2 y - 1) \] 4. **Factor out the common binomial:** \[ (11 y + 1)(2 y - 1) \] **Final Factored Form:** \[ 22 y^{2} - 9 y - 1 = (11 y + 1)(2 y - 1) \] --- ### **v. \( 9 a^{2} + 30 a b + 25 b^{2} \)** This is a perfect square trinomial. To factor it, we can express it in the form \( (ma + nb)^2 \). 1. **Identify the squares and the middle term:** - \( (3a)^2 = 9a^{2} \) - \( (5b)^2 = 25b^{2} \) - The middle term \( 30ab \) is twice the product of \( 3a \) and \( 5b \): \[ 2 \times 3a \times 5b = 30ab \] 2. **Write the factored form:** \[ (3a + 5b)^2 \] **Final Factored Form:** \[ 9 a^{2} + 30 a b + 25 b^{2} = (3a + 5b)^2 \] --- If you have any more questions or need further clarification, feel free to ask!