Comprehensive worksheet featuring 12 mixed practice problems for factoring polynomials in an Algebra 2 or Trigonometry class.

Problem Description:

The task involves solving a series of algebraic problems related to factoring polynomials. The worksheet is divided into three sections:

1. GCF (Greatest Common Factor): Factoring out the greatest common factor.
2. Trinomials with \(a = 1\): Factoring trinomials of the form \(x^2 + bx + c\).
3. Higher-degree polynomials: Factoring more complex polynomials, often involving grouping or substitution.

Below, I will solve each problem step by step.

---

Section 1: GCF (Greatest Common Factor)

#### Problem 1: \(25x^4 - 9\)
- Step 1: Identify the GCF of the terms \(25x^4\) and \(-9\).
- The coefficients are \(25\) and \(-9\). Their GCF is \(1\).
- The variable part is \(x^4\), but since \(-9\) has no \(x\), the GCF for the variables is \(1\).
- Therefore, the GCF is \(1\).

- Step 2: Since there is no common factor other than \(1\), this polynomial cannot be factored further using GCF. However, it can be factored as a difference of squares:
\[
25x^4 - 9 = (5x^2)^2 - 3^2
\]
Using the difference of squares formula \(a^2 - b^2 = (a - b)(a + b)\):
\[
25x^4 - 9 = (5x^2 - 3)(5x^2 + 3)
\]

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

---

#### Problem 4: \(x^3 - x^2 - 4x - 4\)
- Step 1: Group the terms to look for common factors:
\[
x^3 - x^2 - 4x - 4 = (x^3 - x^2) + (-4x - 4)
\]
- Step 2: Factor out the GCF from each group:
\[
x^3 - x^2 = x^2(x - 1)
\]
\[
-4x - 4 = -4(x + 1)
\]
So, the expression becomes:
\[
x^2(x - 1) - 4(x + 1)
\]

- Step 3: Notice that there is no further common factor between the two groups. This polynomial does not factor nicely using simple methods. If additional context or techniques (like synthetic division or the Rational Root Theorem) are required, they should be applied.

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

---

#### Problem 7: \(x^3 + 2x^2 + x + 2\)
- Step 1: Group the terms:
\[
x^3 + 2x^2 + x + 2 = (x^3 + 2x^2) + (x + 2)
\]
- Step 2: Factor out the GCF from each group:
\[
x^3 + 2x^2 = x^2(x + 2)
\]
\[
x + 2 = 1(x + 2)
\]
So, the expression becomes:
\[
x^2(x + 2) + 1(x + 2)
\]

- Step 3: Factor out the common binomial factor \((x + 2)\):
\[
x^2(x + 2) + 1(x + 2) = (x + 2)(x^2 + 1)
\]

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

---

#### Problem 10: \(5xy^2 + 25xy - 15x^2y^2\)
- Step 1: Identify the GCF of the terms \(5xy^2\), \(25xy\), and \(-15x^2y^2\).
- The coefficients are \(5\), \(25\), and \(-15\). Their GCF is \(5\).
- The variables are \(xy^2\), \(xy\), and \(x^2y^2\). The lowest powers are \(x\) and \(y\). So, the GCF for the variables is \(xy\).
- Therefore, the GCF is \(5xy\).

- Step 2: Factor out \(5xy\) from each term:
\[
5xy^2 + 25xy - 15x^2y^2 = 5xy(y + 5 - 3xy)
\]

- Final Answer:
\[
\boxed{5xy(y + 5 - 3xy)}
\]

---

Section 2: Trinomials with \(a = 1\)

#### Problem 2: \(2x^2 - 8x + 12\)
- Step 1: Factor out the GCF of the coefficients \(2\), \(-8\), and \(12\). The GCF is \(2\):
\[
2x^2 - 8x + 12 = 2(x^2 - 4x + 6)
\]

- Step 2: Check if the trinomial \(x^2 - 4x + 6\) can be factored further. Look for two numbers that multiply to \(6\) and add to \(-4\). There are no such integers, so \(x^2 - 4x + 6\) is prime.

- Final Answer:
\[
\boxed{2(x^2 - 4x + 6)}
\]

---

#### Problem 5: \(x^2 - 8\)
- Step 1: Recognize that this is a difference of squares:
\[
x^2 - 8 = x^2 - (\sqrt{8})^2 = x^2 - (2\sqrt{2})^2
\]
- Step 2: Apply the difference of squares formula \(a^2 - b^2 = (a - b)(a + b)\):
\[
x^2 - 8 = (x - 2\sqrt{2})(x + 2\sqrt{2})
\]

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

---

#### Problem 8: \(x^4 - 3x^2 + 2\)
- Step 1: Let \(u = x^2\). Then the polynomial becomes:
\[
x^4 - 3x^2 + 2 = u^2 - 3u + 2
\]
- Step 2: Factor the quadratic \(u^2 - 3u + 2\). Look for two numbers that multiply to \(2\) and add to \(-3\). These numbers are \(-1\) and \(-2\):
\[
u^2 - 3u + 2 = (u - 1)(u - 2)
\]
- Step 3: Substitute back \(u = x^2\):
\[
(u - 1)(u - 2) = (x^2 - 1)(x^2 - 2)
\]
- Step 4: Factor \(x^2 - 1\) as a difference of squares:
\[
x^2 - 1 = (x - 1)(x + 1)
\]
So the full factorization is:
\[
(x^2 - 1)(x^2 - 2) = (x - 1)(x + 1)(x^2 - 2)
\]

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

---

#### Problem 11: \(2x^2 - 9x - 10x + 15\)
- Step 1: Combine like terms:
\[
2x^2 - 9x - 10x + 15 = 2x^2 - 19x + 15
\]
- Step 2: Factor the trinomial \(2x^2 - 19x + 15\). Use the AC method:
- Multiply the leading coefficient \(2\) and the constant term \(15\): \(2 \cdot 15 = 30\).
- Find two numbers that multiply to \(30\) and add to \(-19\). These numbers are \(-15\) and \(-4\).
- Rewrite the middle term \(-19x\) as \(-15x - 4x\):
\[
2x^2 - 19x + 15 = 2x^2 - 15x - 4x + 15
\]
- Group the terms:
\[
(2x^2 - 15x) + (-4x + 15)
\]
- Factor out the GCF from each group:
\[
x(2x - 15) - 1(2x - 15)
\]
- Factor out the common binomial factor \((2x - 15)\):
\[
(2x - 15)(x - 1)
\]

- Final Answer:
\[
\boxed{(2x - 15)(x - 1)}
\]

---

Section 3: Higher-Degree Polynomials

#### Problem 3: \(x^3 + 216\)
- Step 1: Recognize that this is a sum of cubes:
\[
x^3 + 216 = x^3 + 6^3
\]
- Step 2: Apply the sum of cubes formula \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\):
\[
x^3 + 6^3 = (x + 6)(x^2 - 6x + 36)
\]

- Final Answer:
\[
\boxed{(x + 6)(x^2 - 6x + 36)}
\]

---

#### Problem 6: \(3x^2 + 2x^2 - 9x + 6\)
- Step 1: Combine like terms:
\[
3x^2 + 2x^2 - 9x + 6 = 5x^2 - 9x + 6
\]
- Step 2: Factor the trinomial \(5x^2 - 9x + 6\). Use the AC method:
- Multiply the leading coefficient \(5\) and the constant term \(6\): \(5 \cdot 6 = 30\).
- Find two numbers that multiply to \(30\) and add to \(-9\). These numbers are \(-6\) and \(-3\).
- Rewrite the middle term \(-9x\) as \(-6x - 3x\):
\[
5x^2 - 9x + 6 = 5x^2 - 6x - 3x + 6
\]
- Group the terms:
\[
(5x^2 - 6x) + (-3x + 6)
\]
- Factor out the GCF from each group:
\[
x(5x - 6) - 1(5x - 6)
\]
- Factor out the common binomial factor \((5x - 6)\):
\[
(5x - 6)(x - 1)
\]

- Final Answer:
\[
\boxed{(5x - 6)(x - 1)}
\]

---

#### Problem 9: \(36x^4 - 9x^2\)
- Step 1: Factor out the GCF of the terms \(36x^4\) and \(-9x^2\). The GCF is \(9x^2\):
\[
36x^4 - 9x^2 = 9x^2(4x^2 - 1)
\]
- Step 2: Recognize that \(4x^2 - 1\) is a difference of squares:
\[
4x^2 - 1 = (2x)^2 - 1^2
\]
- Step 3: Apply the difference of squares formula \(a^2 - b^2 = (a - b)(a + b)\):
\[
4x^2 - 1 = (2x - 1)(2x + 1)
\]
- Step 4: Substitute back:
\[
36x^4 - 9x^2 = 9x^2(2x - 1)(2x + 1)
\]

- Final Answer:
\[
\boxed{9x^2(2x - 1)(2x + 1)}
\]

---

#### Problem 12: \(x^4 - 1\)
- Step 1: Recognize that this is a difference of squares:
\[
x^4 - 1 = (x^2)^2 - 1^2
\]
- Step 2: Apply the difference of squares formula \(a^2 - b^2 = (a - b)(a + b)\):
\[
x^4 - 1 = (x^2 - 1)(x^2 + 1)
\]
- Step 3: Factor \(x^2 - 1\) as a difference of squares:
\[
x^2 - 1 = (x - 1)(x + 1)
\]
- Step 4: Substitute back:
\[
x^4 - 1 = (x - 1)(x + 1)(x^2 + 1)
\]

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

---

Final Answers:

1. \(\boxed{(5x^2 - 3)(5x^2 + 3)}\)
2. \(\boxed{2(x^2 - 4x + 6)}\)
3. \(\boxed{(x + 6)(x^2 - 6x + 36)}\)
4. \(\boxed{x^2(x - 1) - 4(x + 1)}\)
5. \(\boxed{(x - 2\sqrt{2})(x + 2\sqrt{2})}\)
6. \(\boxed{(5x - 6)(x - 1)}\)
7. \(\boxed{(x + 2)(x^2 + 1)}\)
8. \(\boxed{(x - 1)(x + 1)(x^2 - 2)}\)
9. \(\boxed{9x^2(2x - 1)(2x + 1)}\)
10. \(\boxed{5xy(y + 5 - 3xy)}\)
11. \(\boxed{(2x - 15)(x - 1)}\)
12. \(\boxed{(x - 1)(x + 1)(x^2 + 1)}\)