Answer
**31) \( 7n^{3}+21n^{2}-5n-15 = (n+3)(7n^{2}-5) \)**
**33) \( 15ab-2b^{2}-6a+5b^{3} = (5b-2)(3a+b^{2}) \)**
**35) \( 5mn-10+2m-25n = (m-5)(5n+2) \)**
**37) \( 4uv+14u^{2}+12v+42u = 2(u+3)(2v+7u) \)**
**3) \( 56ab+14-49a-16b = (8b-7)(7a-2) \)**
Solution
**31) Factor \( 7 n^{3}+21 n^{2}-5 n-15 \)**
1. Group the terms:
\[
(7n^3 + 21n^2) + (-5n - 15)
\]
2. Factor each group:
\[
7n^2(n+3) - 5(n+3)
\]
3. Factor out the common factor \( (n+3) \):
\[
(n+3)(7n^2 - 5)
\]
---
**33) Factor \( 15ab-2b^{2}-6a+5b^{3} \)**
1. Rearrange the terms:
\[
15ab -6a -2b^2 +5b^3
\]
2. Group the terms:
\[
(15ab - 6a) + (-2b^2 + 5b^3)
\]
3. Factor each group:
- From \( 15ab - 6a \), factor \( 3a \):
\[
3a(5b-2)
\]
- From \( -2b^2 + 5b^3 \), factor \( b^2 \):
\[
b^2(5b-2)
\]
4. Factor out the common binomial \( (5b-2) \):
\[
(5b-2)(3a + b^2)
\]
---
**35) Factor \( 5mn-10+2m-25n \)**
1. Rearrange the terms to group like terms:
\[
5mn - 25n + 2m - 10
\]
2. Group the terms:
\[
(5mn - 25n) + (2m - 10)
\]
3. Factor each group:
- From \( 5mn - 25n \), factor \( 5n \):
\[
5n(m-5)
\]
- From \( 2m - 10 \), factor \( 2 \):
\[
2(m-5)
\]
4. Factor out the common binomial \( (m-5) \):
\[
(m-5)(5n + 2)
\]
---
**37) Factor \( 4uv+14u^{2}+12v+42u \)**
1. Rearrange and group the terms:
\[
(4uv + 12v) + (14u^2 + 42u)
\]
2. Factor each group:
- From \( 4uv + 12v \), factor \( 4v \):
\[
4v(u+3)
\]
- From \( 14u^2 + 42u \), factor \( 14u \):
\[
14u(u+3)
\]
3. Factor out the common binomial \( (u+3) \):
\[
(u+3)(4v + 14u)
\]
4. Factor a common factor of \( 2 \) from the second binomial:
\[
(u+3) \cdot 2(2v + 7u) = 2(u+3)(2v+7u)
\]
---
**3) Factor \( 56ab+14-49a-16b \)**
1. Rearrange the terms:
\[
56ab - 49a - 16b + 14
\]
2. Group the terms:
\[
(56ab - 49a) + (-16b + 14)
\]
3. Factor each group:
- From \( 56ab - 49a \), factor \( 7a \):
\[
7a(8b-7)
\]
- From \( -16b + 14 \), factor \(-2\):
\[
-2(8b-7)
\]
4. Factor out the common factor \( (8b-7) \):
\[
(8b-7)(7a-2)
\]
Answered by UpStudy AI and reviewed by a Professional Tutor

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