Factor Trinomials Show work for all problems algebraically, using factor by grouping \[ \begin{array}{l}\text { 8. } 2 r^{2}+27 r+70 \\ \text { 10. } 7 a^{2}+53 a+28\end{array} \]
Solución de tutoría real
Responder
Solución
Revisado y aprobado por el equipo de tutoría de UpStudy
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.