Practice your algebra skills with this 15-problem worksheet covering GCF and polynomial factoring.

Problem: Factoring Polynomials Worksheet

The task is to factor each polynomial expression. We will solve each problem step by step.

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#### 1. \( 2x^2y - 2xy \)

- Step 1: Identify the greatest common factor (GCF) of the terms.
- The terms are \( 2x^2y \) and \( -2xy \).
- The GCF of the coefficients is \( 2 \).
- The GCF of the variables is \( xy \).
- Therefore, the GCF is \( 2xy \).

- Step 2: Factor out the GCF.
\[
2x^2y - 2xy = 2xy(x - 1)
\]

- Final Answer:
\[
\boxed{2xy(x - 1)}
\]

---

#### 2. \( 6a^2 - 7a - 10 \)

- Step 1: This is a quadratic trinomial of the form \( ax^2 + bx + c \). We need to factor it into two binomials.
- Here, \( a = 6 \), \( b = -7 \), and \( c = -10 \).

- Step 2: Find two numbers that multiply to \( ac \) and add to \( b \).
- \( ac = 6 \cdot (-10) = -60 \)
- We need two numbers that multiply to \( -60 \) and add to \( -7 \).
- The numbers are \( -12 \) and \( 5 \) because \( -12 \cdot 5 = -60 \) and \( -12 + 5 = -7 \).

- Step 3: Rewrite the middle term using these numbers.
\[
6a^2 - 7a - 10 = 6a^2 - 12a + 5a - 10
\]

- Step 4: Factor by grouping.
\[
6a^2 - 12a + 5a - 10 = 6a(a - 2) + 5(a - 2)
\]
\[
= (6a + 5)(a - 2)
\]

- Final Answer:
\[
\boxed{(6a + 5)(a - 2)}
\]

---

#### 3. \( y^3 + 8 \)

- Step 1: Recognize that this is a sum of cubes.
- The sum of cubes formula is:
\[
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
\]
- Here, \( y^3 + 8 = y^3 + 2^3 \), so \( a = y \) and \( b = 2 \).

- Step 2: Apply the sum of cubes formula.
\[
y^3 + 8 = (y + 2)(y^2 - y \cdot 2 + 2^2)
\]
\[
= (y + 2)(y^2 - 2y + 4)
\]

- Final Answer:
\[
\boxed{(y + 2)(y^2 - 2y + 4)}
\]

---

#### 4. \( x^2 - 8x - 9 \)

- Step 1: This is a quadratic trinomial. We need to factor it into two binomials.
- Here, \( a = 1 \), \( b = -8 \), and \( c = -9 \).

- Step 2: Find two numbers that multiply to \( ac \) and add to \( b \).
- \( ac = 1 \cdot (-9) = -9 \)
- We need two numbers that multiply to \( -9 \) and add to \( -8 \).
- The numbers are \( -9 \) and \( 1 \) because \( -9 \cdot 1 = -9 \) and \( -9 + 1 = -8 \).

- Step 3: Rewrite the middle term using these numbers.
\[
x^2 - 8x - 9 = x^2 - 9x + x - 9
\]

- Step 4: Factor by grouping.
\[
x^2 - 9x + x - 9 = x(x - 9) + 1(x - 9)
\]
\[
= (x + 1)(x - 9)
\]

- Final Answer:
\[
\boxed{(x + 1)(x - 9)}
\]

---

#### 5. \( 15x^2 - 18x - 24 \)

- Step 1: Factor out the GCF of the coefficients.
- The GCF of \( 15 \), \( -18 \), and \( -24 \) is \( 3 \).
\[
15x^2 - 18x - 24 = 3(5x^2 - 6x - 8)
\]

- Step 2: Factor the quadratic \( 5x^2 - 6x - 8 \).
- Here, \( a = 5 \), \( b = -6 \), and \( c = -8 \).
- Find two numbers that multiply to \( ac \) and add to \( b \).
- \( ac = 5 \cdot (-8) = -40 \)
- We need two numbers that multiply to \( -40 \) and add to \( -6 \).
- The numbers are \( -10 \) and \( 4 \) because \( -10 \cdot 4 = -40 \) and \( -10 + 4 = -6 \).

- Step 3: Rewrite the middle term using these numbers.
\[
5x^2 - 6x - 8 = 5x^2 - 10x + 4x - 8
\]

- Step 4: Factor by grouping.
\[
5x^2 - 10x + 4x - 8 = 5x(x - 2) + 4(x - 2)
\]
\[
= (5x + 4)(x - 2)
\]

- Step 5: Combine with the GCF.
\[
15x^2 - 18x - 24 = 3(5x + 4)(x - 2)
\]

- Final Answer:
\[
\boxed{3(5x + 4)(x - 2)}
\]

---

#### 6. \( 25x^3 + 8x^2 \)

- Step 1: Factor out the GCF of the terms.
- The GCF of \( 25x^3 \) and \( 8x^2 \) is \( x^2 \).
\[
25x^3 + 8x^2 = x^2(25x + 8)
\]

- Final Answer:
\[
\boxed{x^2(25x + 8)}
\]

---

#### 7. \( 12x^2 - 16x \)

- Step 1: Factor out the GCF of the terms.
- The GCF of \( 12x^2 \) and \( -16x \) is \( 4x \).
\[
12x^2 - 16x = 4x(3x - 4)
\]

- Final Answer:
\[
\boxed{4x(3x - 4)}
\]

---

#### 8. \( 3x^3 - 6x^2 - 9x \)

- Step 1: Factor out the GCF of the terms.
- The GCF of \( 3x^3 \), \( -6x^2 \), and \( -9x \) is \( 3x \).
\[
3x^3 - 6x^2 - 9x = 3x(x^2 - 2x - 3)
\]

- Step 2: Factor the quadratic \( x^2 - 2x - 3 \).
- Here, \( a = 1 \), \( b = -2 \), and \( c = -3 \).
- Find two numbers that multiply to \( ac \) and add to \( b \).
- \( ac = 1 \cdot (-3) = -3 \)
- We need two numbers that multiply to \( -3 \) and add to \( -2 \).
- The numbers are \( -3 \) and \( 1 \) because \( -3 \cdot 1 = -3 \) and \( -3 + 1 = -2 \).

- Step 3: Rewrite the middle term using these numbers.
\[
x^2 - 2x - 3 = x^2 - 3x + x - 3
\]

- Step 4: Factor by grouping.
\[
x^2 - 3x + x - 3 = x(x - 3) + 1(x - 3)
\]
\[
= (x + 1)(x - 3)
\]

- Step 5: Combine with the GCF.
\[
3x^3 - 6x^2 - 9x = 3x(x + 1)(x - 3)
\]

- Final Answer:
\[
\boxed{3x(x + 1)(x - 3)}
\]

---

#### 9. \( x^2 + 9x + 20 \)

- Step 1: This is a quadratic trinomial. We need to factor it into two binomials.
- Here, \( a = 1 \), \( b = 9 \), and \( c = 20 \).

- Step 2: Find two numbers that multiply to \( ac \) and add to \( b \).
- \( ac = 1 \cdot 20 = 20 \)
- We need two numbers that multiply to \( 20 \) and add to \( 9 \).
- The numbers are \( 4 \) and \( 5 \) because \( 4 \cdot 5 = 20 \) and \( 4 + 5 = 9 \).

- Step 3: Rewrite the middle term using these numbers.
\[
x^2 + 9x + 20 = x^2 + 4x + 5x + 20
\]

- Step 4: Factor by grouping.
\[
x^2 + 4x + 5x + 20 = x(x + 4) + 5(x + 4)
\]
\[
= (x + 5)(x + 4)
\]

- Final Answer:
\[
\boxed{(x + 5)(x + 4)}
\]

---

#### 10. \( 6a^2 - 11a + 4 \)

- Step 1: This is a quadratic trinomial. We need to factor it into two binomials.
- Here, \( a = 6 \), \( b = -11 \), and \( c = 4 \).

- Step 2: Find two numbers that multiply to \( ac \) and add to \( b \).
- \( ac = 6 \cdot 4 = 24 \)
- We need two numbers that multiply to \( 24 \) and add to \( -11 \).
- The numbers are \( -8 \) and \( -3 \) because \( -8 \cdot -3 = 24 \) and \( -8 + (-3) = -11 \).

- Step 3: Rewrite the middle term using these numbers.
\[
6a^2 - 11a + 4 = 6a^2 - 8a - 3a + 4
\]

- Step 4: Factor by grouping.
\[
6a^2 - 8a - 3a + 4 = 2a(3a - 4) - 1(3a - 4)
\]
\[
= (2a - 1)(3a - 4)
\]

- Final Answer:
\[
\boxed{(2a - 1)(3a - 4)}
\]

---

#### 11. \( x^2 - 9x + 20 \)

- Step 1: This is a quadratic trinomial. We need to factor it into two binomials.
- Here, \( a = 1 \), \( b = -9 \), and \( c = 20 \).

- Step 2: Find two numbers that multiply to \( ac \) and add to \( b \).
- \( ac = 1 \cdot 20 = 20 \)
- We need two numbers that multiply to \( 20 \) and add to \( -9 \).
- The numbers are \( -4 \) and \( -5 \) because \( -4 \cdot -5 = 20 \) and \( -4 + (-5) = -9 \).

- Step 3: Rewrite the middle term using these numbers.
\[
x^2 - 9x + 20 = x^2 - 4x - 5x + 20
\]

- Step 4: Factor by grouping.
\[
x^2 - 4x - 5x + 20 = x(x - 4) - 5(x - 4)
\]
\[
= (x - 5)(x - 4)
\]

- Final Answer:
\[
\boxed{(x - 5)(x - 4)}
\]

---

#### 12. \( 6x^2 + 13x + 6 \)

- Step 1: This is a quadratic trinomial. We need to factor it into two binomials.
- Here, \( a = 6 \), \( b = 13 \), and \( c = 6 \).

- Step 2: Find two numbers that multiply to \( ac \) and add to \( b \).
- \( ac = 6 \cdot 6 = 36 \)
- We need two numbers that multiply to \( 36 \) and add to \( 13 \).
- The numbers are \( 9 \) and \( 4 \) because \( 9 \cdot 4 = 36 \) and \( 9 + 4 = 13 \).

- Step 3: Rewrite the middle term using these numbers.
\[
6x^2 + 13x + 6 = 6x^2 + 9x + 4x + 6
\]

- Step 4: Factor by grouping.
\[
6x^2 + 9x + 4x + 6 = 3x(2x + 3) + 2(2x + 3)
\]
\[
= (3x + 2)(2x + 3)
\]

- Final Answer:
\[
\boxed{(3x + 2)(2x + 3)}
\]

---

#### 13. \( 8x^2 - 6x - 2 \)

- Step 1: Factor out the GCF of the coefficients.
- The GCF of \( 8 \), \( -6 \), and \( -2 \) is \( 2 \).
\[
8x^2 - 6x - 2 = 2(4x^2 - 3x - 1)
\]

- Step 2: Factor the quadratic \( 4x^2 - 3x - 1 \).
- Here, \( a = 4 \), \( b = -3 \), and \( c = -1 \).
- Find two numbers that multiply to \( ac \) and add to \( b \).
- \( ac = 4 \cdot (-1) = -4 \)
- We need two numbers that multiply to \( -4 \) and add to \( -3 \).
- The numbers are \( -4 \) and \( 1 \) because \( -4 \cdot 1 = -4 \) and \( -4 + 1 = -3 \).

- Step 3: Rewrite the middle term using these numbers.
\[
4x^2 - 3x - 1 = 4x^2 - 4x + x - 1
\]

- Step 4: Factor by grouping.
\[
4x^2 - 4x + x - 1 = 4x(x - 1) + 1(x - 1)
\]
\[
= (4x + 1)(x - 1)
\]

- Step 5: Combine with the GCF.
\[
8x^2 - 6x - 2 = 2(4x + 1)(x - 1)
\]

- Final Answer:
\[
\boxed{2(4x + 1)(x - 1)}
\]

---

#### 14. \( 6n^2 + 12n - 144 \)

- Step 1: Factor out the GCF of the coefficients.
- The GCF of \( 6 \), \( 12 \), and \( -144 \) is \( 6 \).
\[
6n^2 + 12n - 144 = 6(n^2 + 2n - 24)
\]

- Step 2: Factor the quadratic \( n^2 + 2n - 24 \).
- Here, \( a = 1 \), \( b = 2 \), and \( c = -24 \).
- Find two numbers that multiply to \( ac \) and add to \( b \).
- \( ac = 1 \cdot (-24) = -24 \)
- We need two numbers that multiply to \( -24 \) and add to \( 2 \).
- The numbers are \( 6 \) and \( -4 \) because \( 6 \cdot -4 = -24 \) and \( 6 + (-4) = 2 \).

- Step 3: Rewrite the middle term using these numbers.
\[
n^2 + 2n - 24 = n^2 + 6n - 4n - 24
\]

- Step 4: Factor by grouping.
\[
n^2 + 6n - 4n - 24 = n(n + 6) - 4(n + 6)
\]
\[
= (n - 4)(n + 6)
\]

- Step 5: Combine with the GCF.
\[
6n^2 + 12n - 144 = 6(n - 4)(n + 6)
\]

- Final Answer:
\[
\boxed{6(n - 4)(n + 6)}
\]

---

#### 15. \( 10x^2 - 105x + 135 \)

- Step 1: Factor out the GCF of the coefficients.
- The GCF of \( 10 \), \( -105 \), and \( 135 \) is \( 5 \).
\[
10x^2 - 105x + 135 = 5(2x^2 - 21x + 27)
\]

- Step 2: Factor the quadratic \( 2x^2 - 21x + 27 \).
- Here, \( a = 2 \), \( b = -21 \), and \( c = 27 \).
- Find two numbers that multiply to \( ac \) and add to \( b \).
- \( ac = 2 \cdot 27 = 54 \)
- We need two numbers that multiply to \( 54 \) and add to \( -21 \).
- The numbers are \( -18 \) and \( -3 \) because \( -18 \cdot -3 = 54 \) and \( -18 + (-3) = -21 \).

- Step 3: Rewrite the middle term using these numbers.
\[
2x^2 - 21x + 27 = 2x^2 - 18x - 3x + 27
\]

- Step 4: Factor by grouping.
\[
2x^2 - 18x - 3x + 27 = 2x(x - 9) - 3(x - 9)
\]
\[
= (2x - 3)(x - 9)
\]

- Step 5: Combine with the GCF.
\[
10x^2 - 105x + 135 = 5(2x - 3)(x - 9)
\]

- Final Answer:
\[
\boxed{5(2x - 3)(x - 9)}
\]

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Final Answers:

1. \( \boxed{2xy(x - 1)} \)
2. \( \boxed{(6a + 5)(a - 2)} \)
3. \( \boxed{(y + 2)(y^2 - 2y + 4)} \)
4. \( \boxed{(x + 1)(x - 9)} \)
5. \( \boxed{3(5x + 4)(x - 2)} \)
6. \( \boxed{x^2(25x + 8)} \)
7. \( \boxed{4x(3x - 4)} \)
8. \( \boxed{3x(x + 1)(x - 3)} \)
9. \( \boxed{(x + 5)(x + 4)} \)
10. \( \boxed{(2a - 1)(3a - 4)} \)
11. \( \boxed{(x - 5)(x - 4)} \)
12. \( \boxed{(3x + 2)(2x + 3)} \)
13. \( \boxed{2(4x + 1)(x - 1)} \)
14. \( \boxed{6(n - 4)(n + 6)} \)
15. \( \boxed{5(2x - 3)(x - 9)} \)