Solving Quadratic Equations by Factoring worksheet - Free Printable

To solve the given quadratic equations, we will use various methods such as factoring, completing the square, or the quadratic formula. Let's solve each equation step by step.

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

#### Step 1: Simplify the equation
Factor out the greatest common factor (GCF), which is 2:
\[
2(x^2 - x - 12) = 0
\]
Divide through by 2:
\[
x^2 - x - 12 = 0
\]

#### Step 2: Factor the quadratic
We need two numbers that multiply to \(-12\) and add to \(-1\). These numbers are \(-4\) and \(3\):
\[
x^2 - x - 12 = (x - 4)(x + 3) = 0
\]

#### Step 3: Solve for \(x\)
Set each factor equal to zero:
\[
x - 4 = 0 \quad \text{or} \quad x + 3 = 0
\]
\[
x = 4 \quad \text{or} \quad x = -3
\]

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

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

#### Step 1: Factor the quadratic
We need two numbers that multiply to \(16\) and add to \(10\). These numbers are \(8\) and \(2\):
\[
x^2 + 10x + 16 = (x + 8)(x + 2) = 0
\]

#### Step 2: Solve for \(x\)
Set each factor equal to zero:
\[
x + 8 = 0 \quad \text{or} \quad x + 2 = 0
\]
\[
x = -8 \quad \text{or} \quad x = -2
\]

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

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

#### Step 1: Factor the quadratic
We need two numbers that multiply to \(4 \cdot 9 = 36\) and add to \(15\). These numbers are \(12\) and \(3\). Rewrite the middle term using these numbers:
\[
4x^2 + 12x + 3x + 9 = 0
\]
Factor by grouping:
\[
4x(x + 3) + 3(x + 3) = 0
\]
\[
(4x + 3)(x + 3) = 0
\]

#### Step 2: Solve for \(x\)
Set each factor equal to zero:
\[
4x + 3 = 0 \quad \text{or} \quad x + 3 = 0
\]
\[
x = -\frac{3}{4} \quad \text{or} \quad x = -3
\]

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

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

#### Step 1: Factor the quadratic
We need two numbers that multiply to \(42\) and add to \(-13\). These numbers are \(-7\) and \(-6\):
\[
x^2 - 13x + 42 = (x - 7)(x - 6) = 0
\]

#### Step 2: Solve for \(x\)
Set each factor equal to zero:
\[
x - 7 = 0 \quad \text{or} \quad x - 6 = 0
\]
\[
x = 7 \quad \text{or} \quad x = 6
\]

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

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

#### Step 1: Use the quadratic formula
The quadratic formula is:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \(a = 6\), \(b = -1\), and \(c = -40\). Substitute these values:
\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(6)(-40)}}{2(6)}
\]
\[
x = \frac{1 \pm \sqrt{1 + 960}}{12}
\]
\[
x = \frac{1 \pm \sqrt{961}}{12}
\]
\[
x = \frac{1 \pm 31}{12}
\]

#### Step 2: Solve for \(x\)
\[
x = \frac{1 + 31}{12} = \frac{32}{12} = \frac{8}{3}
\]
\[
x = \frac{1 - 31}{12} = \frac{-30}{12} = -\frac{5}{2}
\]

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

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

#### Step 1: Factor the quadratic
We need two numbers that multiply to \(2 \cdot 15 = 30\) and add to \(13\). These numbers are \(10\) and \(3\). Rewrite the middle term using these numbers:
\[
2x^2 + 10x + 3x + 15 = 0
\]
Factor by grouping:
\[
2x(x + 5) + 3(x + 5) = 0
\]
\[
(2x + 3)(x + 5) = 0
\]

#### Step 2: Solve for \(x\)
Set each factor equal to zero:
\[
2x + 3 = 0 \quad \text{or} \quad x + 5 = 0
\]
\[
x = -\frac{3}{2} \quad \text{or} \quad x = -5
\]

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

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Final Answers:

1. \( 2x^2 - 2x - 24 = 0 \): \(\boxed{x = 4 \text{ or } x = -3}\)
2. \( x^2 + 10x + 16 = 0 \): \(\boxed{x = -8 \text{ or } x = -2}\)
3. \( 4x^2 + 15x + 9 = 0 \): \(\boxed{x = -\frac{3}{4} \text{ or } x = -3}\)
4. \( x^2 - 13x + 42 = 0 \): \(\boxed{x = 7 \text{ or } x = 6}\)
5. \( 6x^2 - x - 40 = 0 \): \(\boxed{x = \frac{8}{3} \text{ or } x = -\frac{5}{2}}\)
6. \( 2x^2 + 13x + 15 = 0 \): \(\boxed{x = -\frac{3}{2} \text{ or } x = -5}\)