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Factor Trinomials Show work for all problems algebraically, using factor by grouping \[ \begin{array}{l}\text { 8. } 2 r^{2}+27 r+70 \\ \text { 10. } 7 a^{2}+53 a+28\end{array} \]

Ask by Chandler Munoz. in the United States
Feb 05,2025

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**Factored Forms:** 1. \( 2r^{2} + 27r + 70 = (r + 10)(2r + 7) \) 2. \( 7a^{2} + 53a + 28 = (a + 7)(7a + 4) \)

Solución

Sure! Let's factor each trinomial step-by-step using **factor by grouping**. --- ### **Problem 8:** Factor \( 2r^{2} + 27r + 70 \) **Step 1: Identify the coefficients** \[ a = 2, \quad b = 27, \quad c = 70 \] **Step 2: Compute \( a \times c \)** \[ a \times c = 2 \times 70 = 140 \] **Step 3: Find two numbers that multiply to \( 140 \) and add up to \( 27 \)** We need two numbers, \( m \) and \( n \), such that: \[ m \times n = 140 \quad \text{and} \quad m + n = 27 \] After checking factor pairs of \( 140 \): \[ 7 \times 20 = 140 \quad \text{and} \quad 7 + 20 = 27 \] So, \( m = 7 \) and \( n = 20 \). **Step 4: Rewrite the middle term using \( m \) and \( n \)** \[ 2r^{2} + 7r + 20r + 70 \] **Step 5: Group the terms** \[ (2r^{2} + 7r) + (20r + 70) \] **Step 6: Factor out the greatest common factor (GCF) from each group** \[ r(2r + 7) + 10(2r + 7) \] **Step 7: Factor out the common binomial factor \( (2r + 7) \)** \[ (r + 10)(2r + 7) \] **Final Factored Form:** \[ 2r^{2} + 27r + 70 = (r + 10)(2r + 7) \] --- ### **Problem 10:** Factor \( 7a^{2} + 53a + 28 \) **Step 1: Identify the coefficients** \[ a = 7, \quad b = 53, \quad c = 28 \] **Step 2: Compute \( a \times c \)** \[ a \times c = 7 \times 28 = 196 \] **Step 3: Find two numbers that multiply to \( 196 \) and add up to \( 53 \)** We need two numbers, \( m \) and \( n \), such that: \[ m \times n = 196 \quad \text{and} \quad m + n = 53 \] After checking factor pairs of \( 196 \): \[ 4 \times 49 = 196 \quad \text{and} \quad 4 + 49 = 53 \] So, \( m = 4 \) and \( n = 49 \). **Step 4: Rewrite the middle term using \( m \) and \( n \)** \[ 7a^{2} + 4a + 49a + 28 \] **Step 5: Group the terms** \[ (7a^{2} + 4a) + (49a + 28) \] **Step 6: Factor out the greatest common factor (GCF) from each group** \[ a(7a + 4) + 7(7a + 4) \] **Step 7: Factor out the common binomial factor \( (7a + 4) \)** \[ (a + 7)(7a + 4) \] **Final Factored Form:** \[ 7a^{2} + 53a + 28 = (a + 7)(7a + 4) \] --- **Summary of Factored Forms:** 1. \( 2r^{2} + 27r + 70 = (r + 10)(2r + 7) \) 2. \( 7a^{2} + 53a + 28 = (a + 7)(7a + 4) \)

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The Deep Dive

To factor the given trinomials using factor by grouping, we will follow a step-by-step approach for each example. **8. \( 2r^2 + 27r + 70 \)** 1. **Multiply the leading coefficient (2) by the constant term (70):** \[ 2 \times 70 = 140 \] 2. **We need to find two numbers that multiply to 140 and add to 27.** After testing several pairs, we find: \[ 7 \text{ and } 20 \quad (7 \times 20 = 140 \text{ and } 7 + 20 = 27) \] 3. **Rewrite the middle term (27r) using these numbers:** \[ 2r^2 + 7r + 20r + 70 \] 4. **Group the terms:** \[ (2r^2 + 7r) + (20r + 70) \] 5. **Factor out the common factors in each group:** \[ r(2r + 7) + 10(2r + 7) \] 6. **Factor by grouping:** \[ (2r + 7)(r + 10) \] **Final factorization:** \( (2r + 7)(r + 10) \) --- **10. \( 7a^2 + 53a + 28 \)** 1. **Multiply the leading coefficient (7) by the constant term (28):** \[ 7 \times 28 = 196 \] 2. **Find two numbers that multiply to 196 and add to 53.** The numbers we find are: \[ 49 \text{ and } 4 \quad (49 \times 4 = 196 \text{ and } 49 + 4 = 53) \] 3. **Rewrite the middle term (53a) using these numbers:** \[ 7a^2 + 49a + 4a + 28 \] 4. **Group the terms:** \[ (7a^2 + 49a) + (4a + 28) \] 5. **Factor out the common factors in each group:** \[ 7a(a + 7) + 4(a + 7) \] 6. **Factor by grouping:** \[ (a + 7)(7a + 4) \] **Final factorization:** \( (a + 7)(7a + 4) \) And there you have it! Both trinomials are successfully factored using the method of grouping.

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