Worksheet for solving quadratic equations by factoring.

The image you provided contains a worksheet titled "Solving Quadratic Equations By Factoring." The task involves solving quadratic equations using the factoring method. Below, I will solve each problem step by step and explain the process.

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Problem 1: \( x^2 - 25x + 144 = 0 \)

#### Step 1: Identify the coefficients
The quadratic equation is in the standard form \( ax^2 + bx + c = 0 \), where:
- \( a = 1 \)
- \( b = -25 \)
- \( c = 144 \)

#### Step 2: Factor the quadratic expression
We need to find two numbers that multiply to \( ac = 1 \cdot 144 = 144 \) and add up to \( b = -25 \).

The two numbers are -9 and -16 because:
- \( -9 \cdot -16 = 144 \)
- \( -9 + (-16) = -25 \)

#### Step 3: Rewrite the middle term using these numbers
Rewrite \( -25x \) as \( -9x - 16x \):
\[ x^2 - 25x + 144 = x^2 - 9x - 16x + 144 \]

#### Step 4: Factor by grouping
Group the terms:
\[ (x^2 - 9x) + (-16x + 144) \]

Factor out the greatest common factor (GCF) from each group:
\[ x(x - 9) - 16(x - 9) \]

Factor out the common binomial factor \( (x - 9) \):
\[ (x - 9)(x - 16) = 0 \]

#### Step 5: Solve for \( x \)
Set each factor equal to zero:
\[ x - 9 = 0 \quad \text{or} \quad x - 16 = 0 \]
\[ x = 9 \quad \text{or} \quad x = 16 \]

#### Final Answer:
\[ \boxed{x = 9 \text{ or } x = 16} \]

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Problem 2: \( 6x^2 + 5x - 6 = 0 \)

#### Step 1: Identify the coefficients
The quadratic equation is in the standard form \( ax^2 + bx + c = 0 \), where:
- \( a = 6 \)
- \( b = 5 \)
- \( c = -6 \)

#### Step 2: Factor the quadratic expression
We need to find two numbers that multiply to \( ac = 6 \cdot -6 = -36 \) and add up to \( b = 5 \).

The two numbers are 9 and -4 because:
- \( 9 \cdot -4 = -36 \)
- \( 9 + (-4) = 5 \)

#### Step 3: Rewrite the middle term using these numbers
Rewrite \( 5x \) as \( 9x - 4x \):
\[ 6x^2 + 5x - 6 = 6x^2 + 9x - 4x - 6 \]

#### Step 4: Factor by grouping
Group the terms:
\[ (6x^2 + 9x) + (-4x - 6) \]

Factor out the GCF from each group:
\[ 3x(2x + 3) - 2(2x + 3) \]

Factor out the common binomial factor \( (2x + 3) \):
\[ (3x - 2)(2x + 3) = 0 \]

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

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

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Problem 3: \( 3x^2 - 18x + 27 = 0 \)

#### Step 1: Simplify the equation
Factor out the greatest common factor (GCF) of the coefficients, which is 3:
\[ 3(x^2 - 6x + 9) = 0 \]

#### Step 2: Factor the quadratic expression inside the parentheses
The expression \( x^2 - 6x + 9 \) is a perfect square trinomial:
\[ x^2 - 6x + 9 = (x - 3)^2 \]

So the equation becomes:
\[ 3(x - 3)^2 = 0 \]

#### Step 3: Solve for \( x \)
Set the factor equal to zero:
\[ (x - 3)^2 = 0 \]
\[ x - 3 = 0 \]
\[ x = 3 \]

#### Final Answer:
\[ \boxed{x = 3} \]

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Problem 4: \( 4x^2 + 20x + 25 = 0 \)

#### Step 1: Identify the coefficients
The quadratic equation is in the standard form \( ax^2 + bx + c = 0 \), where:
- \( a = 4 \)
- \( b = 20 \)
- \( c = 25 \)

#### Step 2: Factor the quadratic expression
The expression \( 4x^2 + 20x + 25 \) is a perfect square trinomial:
\[ 4x^2 + 20x + 25 = (2x + 5)^2 \]

So the equation becomes:
\[ (2x + 5)^2 = 0 \]

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

#### Final Answer:
\[ \boxed{x = -\frac{5}{2}} \]

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Problem 5: \( 3x^2 - 75 = 0 \)

#### Step 1: Simplify the equation
Factor out the GCF of the coefficients, which is 3:
\[ 3(x^2 - 25) = 0 \]

#### Step 2: Factor the difference of squares
The expression \( x^2 - 25 \) is a difference of squares:
\[ x^2 - 25 = (x - 5)(x + 5) \]

So the equation becomes:
\[ 3(x - 5)(x + 5) = 0 \]

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

#### Final Answer:
\[ \boxed{x = 5 \text{ or } x = -5} \]

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Problem 6: \( 2x^2 - 18 = 0 \)

#### Step 1: Simplify the equation
Factor out the GCF of the coefficients, which is 2:
\[ 2(x^2 - 9) = 0 \]

#### Step 2: Factor the difference of squares
The expression \( x^2 - 9 \) is a difference of squares:
\[ x^2 - 9 = (x - 3)(x + 3) \]

So the equation becomes:
\[ 2(x - 3)(x + 3) = 0 \]

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

#### Final Answer:
\[ \boxed{x = 3 \text{ or } x = -3} \]

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Problem 7: \( 7x^2 - 42x = 0 \)

#### Step 1: Factor out the GCF
The GCF of the terms is \( 7x \):
\[ 7x(x - 6) = 0 \]

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

#### Final Answer:
\[ \boxed{x = 0 \text{ or } x = 6} \]

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Problem 8: \( x^2 - 64 = 0 \)

#### Step 1: Factor the difference of squares
The expression \( x^2 - 64 \) is a difference of squares:
\[ x^2 - 64 = (x - 8)(x + 8) \]

So the equation becomes:
\[ (x - 8)(x + 8) = 0 \]

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

#### Final Answer:
\[ \boxed{x = 8 \text{ or } x = -8} \]

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Problem 9: \( 2x^2 + 12x + 18 = 0 \)

#### Step 1: Simplify the equation
Factor out the GCF of the coefficients, which is 2:
\[ 2(x^2 + 6x + 9) = 0 \]

#### Step 2: Factor the quadratic expression inside the parentheses
The expression \( x^2 + 6x + 9 \) is a perfect square trinomial:
\[ x^2 + 6x + 9 = (x + 3)^2 \]

So the equation becomes:
\[ 2(x + 3)^2 = 0 \]

#### Step 3: Solve for \( x \)
Set the factor equal to zero:
\[ (x + 3)^2 = 0 \]
\[ x + 3 = 0 \]
\[ x = -3 \]

#### Final Answer:
\[ \boxed{x = -3} \]

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Problem 10: \( 3x^2 - 12x = 0 \)

#### Step 1: Factor out the GCF
The GCF of the terms is \( 3x \):
\[ 3x(x - 4) = 0 \]

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

#### Final Answer:
\[ \boxed{x = 0 \text{ or } x = 4} \]

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Final Answers Summary

1. \( \boxed{x = 9 \text{ or } x = 16} \)
2. \( \boxed{x = \frac{2}{3} \text{ or } x = -\frac{3}{2}} \)
3. \( \boxed{x = 3} \)
4. \( \boxed{x = -\frac{5}{2}} \)
5. \( \boxed{x = 5 \text{ or } x = -5} \)
6. \( \boxed{x = 3 \text{ or } x = -3} \)
7. \( \boxed{x = 0 \text{ or } x = 6} \)
8. \( \boxed{x = 8 \text{ or } x = -8} \)
9. \( \boxed{x = -3} \)
10. \( \boxed{x = 0 \text{ or } x = 4} \)

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If you have any further questions or need additional clarification, feel free to ask!