Solving Quadratic Equations by Factoring | - Free Printable

To solve the given quadratic equations, we will use factoring where possible. Let's go through each equation step by step.

---

1. Solve \( 2x^2 - 2x - 24 = 0 \)

#### Step 1: Factor out the greatest common factor (GCF).
The GCF of the coefficients \(2\), \(-2\), and \(-24\) is \(2\). Factor out \(2\):
\[
2(x^2 - x - 12) = 0
\]

#### Step 2: Factor the quadratic expression inside the parentheses.
We need to factor \(x^2 - x - 12\). Find two numbers that multiply to \(-12\) and add to \(-1\). These numbers are \(-4\) and \(3\):
\[
x^2 - x - 12 = (x - 4)(x + 3)
\]

#### Step 3: Write the fully factored equation.
\[
2(x - 4)(x + 3) = 0
\]

#### Step 4: Use the zero-product property.
Set each factor equal to zero:
\[
2 = 0 \quad \text{(not possible)}, \quad x - 4 = 0, \quad x + 3 = 0
\]
Solve for \(x\):
\[
x - 4 = 0 \implies x = 4
\]
\[
x + 3 = 0 \implies x = -3
\]

#### Solution:
\[
x = 4 \quad \text{or} \quad x = -3
\]

---

2. Solve \( x^2 + 10x + 16 = 0 \)

#### Step 1: Factor the quadratic expression.
We need to factor \(x^2 + 10x + 16\). Find two numbers that multiply to \(16\) and add to \(10\). These numbers are \(8\) and \(2\):
\[
x^2 + 10x + 16 = (x + 8)(x + 2)
\]

#### Step 2: Write the factored equation.
\[
(x + 8)(x + 2) = 0
\]

#### Step 3: Use the zero-product property.
Set each factor equal to zero:
\[
x + 8 = 0, \quad x + 2 = 0
\]
Solve for \(x\):
\[
x + 8 = 0 \implies x = -8
\]
\[
x + 2 = 0 \implies x = -2
\]

#### Solution:
\[
x = -8 \quad \text{or} \quad x = -2
\]

---

3. Solve \( 4x^2 + 15x + 9 = 0 \)

#### Step 1: Factor the quadratic expression.
We need to factor \(4x^2 + 15x + 9\). Use the "ac method":
- Multiply the leading coefficient (\(4\)) by the constant term (\(9\)): \(4 \cdot 9 = 36\).
- Find two numbers that multiply to \(36\) and add to the middle coefficient (\(15\)). These numbers are \(12\) and \(3\).

#### Step 2: Rewrite the middle term using these numbers.
\[
4x^2 + 15x + 9 = 4x^2 + 12x + 3x + 9
\]

#### Step 3: Factor by grouping.
Group the terms:
\[
(4x^2 + 12x) + (3x + 9)
\]
Factor out the GCF from each group:
\[
4x(x + 3) + 3(x + 3)
\]
Factor out the common binomial factor \((x + 3)\):
\[
(4x + 3)(x + 3) = 0
\]

#### Step 4: Use the zero-product property.
Set each factor equal to zero:
\[
4x + 3 = 0, \quad x + 3 = 0
\]
Solve for \(x\):
\[
4x + 3 = 0 \implies 4x = -3 \implies x = -\frac{3}{4}
\]
\[
x + 3 = 0 \implies x = -3
\]

#### Solution:
\[
x = -\frac{3}{4} \quad \text{or} \quad x = -3
\]

---

4. Solve \( x^2 - 13x + 42 = 0 \)

#### Step 1: Factor the quadratic expression.
We need to factor \(x^2 - 13x + 42\). Find two numbers that multiply to \(42\) and add to \(-13\). These numbers are \(-7\) and \(-6\):
\[
x^2 - 13x + 42 = (x - 7)(x - 6)
\]

#### Step 2: Write the factored equation.
\[
(x - 7)(x - 6) = 0
\]

#### Step 3: Use the zero-product property.
Set each factor equal to zero:
\[
x - 7 = 0, \quad x - 6 = 0
\]
Solve for \(x\):
\[
x - 7 = 0 \implies x = 7
\]
\[
x - 6 = 0 \implies x = 6
\]

#### Solution:
\[
x = 7 \quad \text{or} \quad x = 6
\]

---

5. Solve \( 6x^2 - x - 40 = 0 \)

#### Step 1: Factor the quadratic expression.
We need to factor \(6x^2 - x - 40\). Use the "ac method":
- Multiply the leading coefficient (\(6\)) by the constant term (\(-40\)): \(6 \cdot (-40) = -240\).
- Find two numbers that multiply to \(-240\) and add to the middle coefficient (\(-1\)). These numbers are \(-16\) and \(15\).

#### Step 2: Rewrite the middle term using these numbers.
\[
6x^2 - x - 40 = 6x^2 - 16x + 15x - 40
\]

#### Step 3: Factor by grouping.
Group the terms:
\[
(6x^2 - 16x) + (15x - 40)
\]
Factor out the GCF from each group:
\[
2x(3x - 8) + 5(3x - 8)
\]
Factor out the common binomial factor \((3x - 8)\):
\[
(2x + 5)(3x - 8) = 0
\]

#### Step 4: Use the zero-product property.
Set each factor equal to zero:
\[
2x + 5 = 0, \quad 3x - 8 = 0
\]
Solve for \(x\):
\[
2x + 5 = 0 \implies 2x = -5 \implies x = -\frac{5}{2}
\]
\[
3x - 8 = 0 \implies 3x = 8 \implies x = \frac{8}{3}
\]

#### Solution:
\[
x = -\frac{5}{2} \quad \text{or} \quad x = \frac{8}{3}
\]

---

6. Solve \( 2x^2 + 13x + 15 = 0 \)

#### Step 1: Factor the quadratic expression.
We need to factor \(2x^2 + 13x + 15\). Use the "ac method":
- Multiply the leading coefficient (\(2\)) by the constant term (\(15\)): \(2 \cdot 15 = 30\).
- Find two numbers that multiply to \(30\) and add to the middle coefficient (\(13\)). These numbers are \(10\) and \(3\).

#### Step 2: Rewrite the middle term using these numbers.
\[
2x^2 + 13x + 15 = 2x^2 + 10x + 3x + 15
\]

#### Step 3: Factor by grouping.
Group the terms:
\[
(2x^2 + 10x) + (3x + 15)
\]
Factor out the GCF from each group:
\[
2x(x + 5) + 3(x + 5)
\]
Factor out the common binomial factor \((x + 5)\):
\[
(2x + 3)(x + 5) = 0
\]

#### Step 4: Use the zero-product property.
Set each factor equal to zero:
\[
2x + 3 = 0, \quad x + 5 = 0
\]
Solve for \(x\):
\[
2x + 3 = 0 \implies 2x = -3 \implies x = -\frac{3}{2}
\]
\[
x + 5 = 0 \implies x = -5
\]

#### Solution:
\[
x = -\frac{3}{2} \quad \text{or} \quad x = -5
\]

---

Final Answers:

\[
\boxed{
\begin{aligned}
1. & \quad x = 4 \quad \text{or} \quad x = -3 \\
2. & \quad x = -8 \quad \text{or} \quad x = -2 \\
3. & \quad x = -\frac{3}{4} \quad \text{or} \quad x = -3 \\
4. & \quad x = 7 \quad \text{or} \quad x = 6 \\
5. & \quad x = -\frac{5}{2} \quad \text{or} \quad x = \frac{8}{3} \\
6. & \quad x = -\frac{3}{2} \quad \text{or} \quad x = -5
\end{aligned}
}
\]