Solving Quadratic Equations by Factoring worksheet with problems and solutions.

Problem: Solving Quadratic Equations by Factoring

The task involves solving quadratic equations by factoring. Each equation is provided, and the solutions are already listed in the image. Below, I will explain the process of solving each equation step by step.

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General Approach:

1. Set the equation to zero: Ensure the quadratic equation is in the form \( ax^2 + bx + c = 0 \).
2. Factor the quadratic expression: Break down the quadratic into two binomials.
3. Use the Zero Product Property: If \( (x - p)(x - q) = 0 \), then \( x - p = 0 \) or \( x - q = 0 \).
4. Solve for \( x \): Find the values of \( x \) that satisfy the equation.

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Solutions:

#### 1. \( x(x + 16 - 1) = 0 \)
- Simplify inside the parentheses: \( x(x + 15) = 0 \).
- Factor: The equation is already factored as \( x(x + 15) = 0 \).
- Use the Zero Product Property:
- \( x = 0 \)
- \( x + 15 = 0 \implies x = -15 \)
- Solutions: \( \{0, -15\} \)

#### 2. \( 2x^2 + 8x + 2 = 0 \)
- Factor out the greatest common factor (GCF): \( 2(x^2 + 4x + 1) = 0 \).
- Focus on factoring \( x^2 + 4x + 1 \). However, this does not factor nicely over the integers. Instead, use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 1 \), \( b = 4 \), \( c = 1 \):
\[
x = \frac{-4 \pm \sqrt{4^2 - 4(1)(1)}}{2(1)} = \frac{-4 \pm \sqrt{16 - 4}}{2} = \frac{-4 \pm \sqrt{12}}{2} = \frac{-4 \pm 2\sqrt{3}}{2} = -2 \pm \sqrt{3}
\]
- Solutions: \( \{-2 + \sqrt{3}, -2 - \sqrt{3}\} \)

#### 3. \( 54x + 54x + 1 = 0 \)
- Combine like terms: \( 108x + 1 = 0 \).
- This is a linear equation, not a quadratic equation. Solve for \( x \):
\[
108x = -1 \implies x = -\frac{1}{108}
\]
- Solutions: \( \left\{-\frac{1}{108}\right\} \)

#### 4. \( 12m = 2(4m + 2) \)
- Distribute on the right-hand side: \( 12m = 8m + 4 \).
- Move all terms to one side: \( 12m - 8m - 4 = 0 \).
- Simplify: \( 4m - 4 = 0 \).
- Solve for \( m \):
\[
4m = 4 \implies m = 1
\]
- Solutions: \( \{1\} \)

#### 5. \( x^2 - 11x + 19 = -5 \)
- Move all terms to one side: \( x^2 - 11x + 19 + 5 = 0 \).
- Simplify: \( x^2 - 11x + 24 = 0 \).
- Factor: \( (x - 3)(x - 8) = 0 \).
- Use the Zero Product Property:
- \( x - 3 = 0 \implies x = 3 \)
- \( x - 8 = 0 \implies x = 8 \)
- Solutions: \( \{3, 8\} \)

#### 6. \( n^2 - 10n + 15 = 5 \)
- Move all terms to one side: \( n^2 - 10n + 15 - 5 = 0 \).
- Simplify: \( n^2 - 10n + 10 = 0 \).
- This does not factor nicely over the integers. Use the quadratic formula:
\[
n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 1 \), \( b = -10 \), \( c = 10 \):
\[
n = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(10)}}{2(1)} = \frac{10 \pm \sqrt{100 - 40}}{2} = \frac{10 \pm \sqrt{60}}{2} = \frac{10 \pm 2\sqrt{15}}{2} = 5 \pm \sqrt{15}
\]
- Solutions: \( \{5 + \sqrt{15}, 5 - \sqrt{15}\} \)

#### 7. \( u^2 - 10u + 22 = -2 \)
- Move all terms to one side: \( u^2 - 10u + 22 + 2 = 0 \).
- Simplify: \( u^2 - 10u + 24 = 0 \).
- Factor: \( (u - 6)(u - 4) = 0 \).
- Use the Zero Product Property:
- \( u - 6 = 0 \implies u = 6 \)
- \( u - 4 = 0 \implies u = 4 \)
- Solutions: \( \{6, 4\} \)

#### 8. \( x^2 + 3x - 12 = 6 \)
- Move all terms to one side: \( x^2 + 3x - 12 - 6 = 0 \).
- Simplify: \( x^2 + 3x - 18 = 0 \).
- Factor: \( (x + 6)(x - 3) = 0 \).
- Use the Zero Product Property:
- \( x + 6 = 0 \implies x = -6 \)
- \( x - 3 = 0 \implies x = 3 \)
- Solutions: \( \{3, -6\} \)

#### 9. \( 6a^2 - 18a + 18 = 6 \)
- Move all terms to one side: \( 6a^2 - 18a + 18 - 6 = 0 \).
- Simplify: \( 6a^2 - 18a + 12 = 0 \).
- Factor out the GCF: \( 6(a^2 - 3a + 2) = 0 \).
- Factor the quadratic: \( a^2 - 3a + 2 = (a - 1)(a - 2) \).
- Use the Zero Product Property:
- \( a - 1 = 0 \implies a = 1 \)
- \( a - 2 = 0 \implies a = 2 \)
- Solutions: \( \{1, 2\} \)

#### 10. \( 7v^2 - 4v = -7 \)
- Move all terms to one side: \( 7v^2 - 4v + 7 = 0 \).
- This does not factor nicely over the integers. Use the quadratic formula:
\[
v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 7 \), \( b = -4 \), \( c = 7 \):
\[
v = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(7)(7)}}{2(7)} = \frac{4 \pm \sqrt{16 - 196}}{14} = \frac{4 \pm \sqrt{-180}}{14} = \frac{4 \pm 6i\sqrt{5}}{14} = \frac{2 \pm 3i\sqrt{5}}{7}
\]
- Solutions: \( \left\{\frac{2 + 3i\sqrt{5}}{7}, \frac{2 - 3i\sqrt{5}}{7}\right\} \)

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Final Answer:

\[
\boxed{
\begin{aligned}
1. & \{0, -15\} \\
2. & \{-2 + \sqrt{3}, -2 - \sqrt{3}\} \\
3. & \left\{-\frac{1}{108}\right\} \\
4. & \{1\} \\
5. & \{3, 8\} \\
6. & \{5 + \sqrt{15}, 5 - \sqrt{15}\} \\
7. & \{6, 4\} \\
8. & \{3, -6\} \\
9. & \{1, 2\} \\
10. & \left\{\frac{2 + 3i\sqrt{5}}{7}, \frac{2 - 3i\sqrt{5}}{7}\right\}
\end{aligned}
}
\]