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

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UpStudy AI · Accepted Answer

Sure, let's factor both expressions you've provided.

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### **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:**
$$
22 y^{2} - 9 y - 1 = (11 y + 1)(2 y - 1)
$$

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### **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:**
$$
9 a^{2} + 30 a b + 25 b^{2} = (3a + 5b)^2
$$

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If you have any more questions or need further clarification, feel free to ask!
$$ 22 y^{2} - 9 y - 1 = (11 y + 1)(2 y - 1) $$ $$ 9 a^{2} + 30 a b + 25 b^{2} = (3a + 5b)^2 $$

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

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