Factoring By Grouping Worksheet - Fill and Sign Printable Template ... - Free Printable

Problem: Factor each of the following expressions completely.

The task involves factoring algebraic expressions. Let's solve each problem step by step.

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#### 1) \( 3x + 3y \)
- Step 1: Identify the greatest common factor (GCF) of the terms.
- The GCF of \( 3x \) and \( 3y \) is \( 3 \).
- Step 2: Factor out the GCF.
\[
3x + 3y = 3(x + y)
\]
- Solution:
\[
\boxed{3(x + y)}
\]

---

#### 2) \( w(x - y) - 3(x - y) \)
- Step 1: Notice that \( (x - y) \) is a common factor in both terms.
- Step 2: Factor out \( (x - y) \).
\[
w(x - y) - 3(x - y) = (x - y)(w - 3)
\]
- Solution:
\[
\boxed{(x - y)(w - 3)}
\]

---

#### 3) \( 2m(x - y) + y(x - y) \)
- Step 1: Notice that \( (x - y) \) is a common factor in both terms.
- Step 2: Factor out \( (x - y) \).
\[
2m(x - y) + y(x - y) = (x - y)(2m + y)
\]
- Solution:
\[
\boxed{(x - y)(2m + y)}
\]

---

#### 4) \( 2p(7q - p) - 3q(7q - p) \)
- Step 1: Notice that \( (7q - p) \) is a common factor in both terms.
- Step 2: Factor out \( (7q - p) \).
\[
2p(7q - p) - 3q(7q - p) = (7q - p)(2p - 3q)
\]
- Solution:
\[
\boxed{(7q - p)(2p - 3q)}
\]

---

#### 5) \( 2m(m + 3) - 3(m + 3) \)
- Step 1: Notice that \( (m + 3) \) is a common factor in both terms.
- Step 2: Factor out \( (m + 3) \).
\[
2m(m + 3) - 3(m + 3) = (m + 3)(2m - 3)
\]
- Solution:
\[
\boxed{(m + 3)(2m - 3)}
\]

---

#### 6) \( 3a^2 + 9ab + 3bc \)
- Step 1: Identify the GCF of all terms.
- The GCF of \( 3a^2 \), \( 9ab \), and \( 3bc \) is \( 3 \).
- Step 2: Factor out the GCF.
\[
3a^2 + 9ab + 3bc = 3(a^2 + 3ab + bc)
\]
- Solution:
\[
\boxed{3(a^2 + 3ab + bc)}
\]

---

#### 7) \( x^2 + xy - 2y \)
- Step 1: This expression does not have a common factor for all terms, so we look for other factoring methods (e.g., grouping or quadratic factoring).
- Step 2: Group the terms to see if factoring by grouping works.
\[
x^2 + xy - 2y = x(x + y) - 2y
\]
However, this does not simplify further into a product of binomials.
- Step 3: Since no further factoring is possible, the expression is already in its simplest form.
- Solution:
\[
\boxed{x^2 + xy - 2y}
\]

---

#### 8) \( x^2 - 2x + xy - 2y \)
- Step 1: Group the terms to see if factoring by grouping works.
\[
x^2 - 2x + xy - 2y = (x^2 - 2x) + (xy - 2y)
\]
- Step 2: Factor out the GCF from each group.
\[
(x^2 - 2x) + (xy - 2y) = x(x - 2) + y(x - 2)
\]
- Step 3: Notice that \( (x - 2) \) is a common factor.
\[
x(x - 2) + y(x - 2) = (x - 2)(x + y)
\]
- Solution:
\[
\boxed{(x - 2)(x + y)}
\]

---

#### 9) \( x^2 - 2xy + 4x - 8y \)
- Step 1: Group the terms to see if factoring by grouping works.
\[
x^2 - 2xy + 4x - 8y = (x^2 - 2xy) + (4x - 8y)
\]
- Step 2: Factor out the GCF from each group.
\[
(x^2 - 2xy) + (4x - 8y) = x(x - 2y) + 4(x - 2y)
\]
- Step 3: Notice that \( (x - 2y) \) is a common factor.
\[
x(x - 2y) + 4(x - 2y) = (x - 2y)(x + 4)
\]
- Solution:
\[
\boxed{(x - 2y)(x + 4)}
\]

---

#### 10) \( p^3 + 2p^2 + 8p + 16 \)
- Step 1: Group the terms to see if factoring by grouping works.
\[
p^3 + 2p^2 + 8p + 16 = (p^3 + 2p^2) + (8p + 16)
\]
- Step 2: Factor out the GCF from each group.
\[
(p^3 + 2p^2) + (8p + 16) = p^2(p + 2) + 8(p + 2)
\]
- Step 3: Notice that \( (p + 2) \) is a common factor.
\[
p^2(p + 2) + 8(p + 2) = (p + 2)(p^2 + 8)
\]
- Solution:
\[
\boxed{(p + 2)(p^2 + 8)}
\]

---

#### 11) \( x^3 + x^2 + 6x + 2 \)
- Step 1: Group the terms to see if factoring by grouping works.
\[
x^3 + x^2 + 6x + 2 = (x^3 + x^2) + (6x + 2)
\]
- Step 2: Factor out the GCF from each group.
\[
(x^3 + x^2) + (6x + 2) = x^2(x + 1) + 2(3x + 1)
\]
However, there is no common factor between \( x^2(x + 1) \) and \( 2(3x + 1) \).
- Step 3: Since no further factoring is possible, the expression is already in its simplest form.
- Solution:
\[
\boxed{x^3 + x^2 + 6x + 2}
\]

---

#### 12) \( p^3 - 2p^2 + mp - 2mp \)
- Step 1: Group the terms to see if factoring by grouping works.
\[
p^3 - 2p^2 + mp - 2mp = (p^3 - 2p^2) + (mp - 2mp)
\]
- Step 2: Factor out the GCF from each group.
\[
(p^3 - 2p^2) + (mp - 2mp) = p^2(p - 2) + m(p - 2)
\]
- Step 3: Notice that \( (p - 2) \) is a common factor.
\[
p^2(p - 2) + m(p - 2) = (p - 2)(p^2 + m)
\]
- Solution:
\[
\boxed{(p - 2)(p^2 + m)}
\]

---

#### 13) \( x^3 - 3x^2 - 6x + 18 \)
- Step 1: Group the terms to see if factoring by grouping works.
\[
x^3 - 3x^2 - 6x + 18 = (x^3 - 3x^2) + (-6x + 18)
\]
- Step 2: Factor out the GCF from each group.
\[
(x^3 - 3x^2) + (-6x + 18) = x^2(x - 3) - 6(x - 3)
\]
- Step 3: Notice that \( (x - 3) \) is a common factor.
\[
x^2(x - 3) - 6(x - 3) = (x - 3)(x^2 - 6)
\]
- Solution:
\[
\boxed{(x - 3)(x^2 - 6)}
\]

---

#### 14) \( 3xy - 2y - 12x + 8 \)
- Step 1: Group the terms to see if factoring by grouping works.
\[
3xy - 2y - 12x + 8 = (3xy - 2y) + (-12x + 8)
\]
- Step 2: Factor out the GCF from each group.
\[
(3xy - 2y) + (-12x + 8) = y(3x - 2) - 4(3x - 2)
\]
- Step 3: Notice that \( (3x - 2) \) is a common factor.
\[
y(3x - 2) - 4(3x - 2) = (3x - 2)(y - 4)
\]
- Solution:
\[
\boxed{(3x - 2)(y - 4)}
\]

---

#### 15) \( 2ab - b - 6 + 12c \)
- Step 1: Group the terms to see if factoring by grouping works.
\[
2ab - b - 6 + 12c = (2ab - b) + (-6 + 12c)
\]
- Step 2: Factor out the GCF from each group.
\[
(2ab - b) + (-6 + 12c) = b(2a - 1) - 6(1 - 2c)
\]
However, there is no common factor between \( b(2a - 1) \) and \( -6(1 - 2c) \).
- Step 3: Since no further factoring is possible, the expression is already in its simplest form.
- Solution:
\[
\boxed{2ab - b - 6 + 12c}
\]

---

Final Answers:

\[
\boxed{
\begin{aligned}
1) & \quad 3(x + y) \\
2) & \quad (x - y)(w - 3) \\
3) & \quad (x - y)(2m + y) \\
4) & \quad (7q - p)(2p - 3q) \\
5) & \quad (m + 3)(2m - 3) \\
6) & \quad 3(a^2 + 3ab + bc) \\
7) & \quad x^2 + xy - 2y \\
8) & \quad (x - 2)(x + y) \\
9) & \quad (x - 2y)(x + 4) \\
10) & \quad (p + 2)(p^2 + 8) \\
11) & \quad x^3 + x^2 + 6x + 2 \\
12) & \quad (p - 2)(p^2 + m) \\
13) & \quad (x - 3)(x^2 - 6) \\
14) & \quad (3x - 2)(y - 4) \\
15) & \quad 2ab - b - 6 + 12c
\end{aligned}
}
\]