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QUESTION 25 If possible, factor the polynomial completely. If a polynomial cannot be factored, state that it is prime. \( 9 a^{2}+30 a+25 \)

Ask by Bernard Johnston. in the United States
Jan 31,2025

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Answer

The polynomial factors completely as (3 a + 5) squared.

Solution

To factor the polynomial \( 9a^{2} + 30a + 25 \) completely, let's follow these steps: 1. **Identify the Structure:** The given polynomial is a quadratic in the form \( ax^2 + bx + c \). 2. **Look for Factoring Patterns:** We can attempt to factor it as \( (ma + n)(pa + q) \). 3. **Determine the Coefficients:** - The product of the first terms should be \( 9a^2 \). Possible pairs are: - \( 3a \times 3a = 9a^2 \) - The product of the last terms should be \( 25 \). Possible pairs are: - \( 5 \times 5 = 25 \) 4. **Verify the Middle Term:** Now, check if the middle terms add up to \( 30a \): \[ (3a \times 5) + (3a \times 5) = 15a + 15a = 30a \] This matches the middle term of the original polynomial. 5. **Write the Factored Form:** Thus, the polynomial factors neatly into: \[ (3a + 5)^2 \] **Final Answer:** The polynomial factors completely as (3 a + 5) squared. **Answer:** The polynomial factors completely as (3 a + 5) squared.

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To factor the polynomial \( 9a^2 + 30a + 25 \), we can look for two numbers that multiply to \( 9 \times 25 = 225 \) and add up to \( 30 \). Those numbers are \( 15 \) and \( 15 \). Thus, we can rewrite the polynomial as \( 9a^2 + 15a + 15a + 25 \). Now, grouping them gives us: \( (9a^2 + 15a) + (15a + 25) \). Factoring each group results in \( 3a(3a + 5) + 5(3a + 5) \). Now we can factor out the common binomial \( (3a + 5) \): The completely factored form is \( (3a + 5)(3a + 5) \) or \( (3a + 5)^2 \), showing that the polynomial is a perfect square trinomial!

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