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Lesson Plan: Solving Quadratic Equations by Factoring

 

Lesson Summary

This lesson introduces students to solving quadratic equations by factoring and applying the Zero Product Property. Students will engage in a hands-on warm-up activity, explore three detailed examples, and practice with a variety of quadratic equations. Multimedia resources from Media4Math.com are incorporated throughout, including vocabulary definitions, clip art, and tutorial videos. The lesson concludes with a 10-question quiz and an answer key. Estimated time: 50 minutes.

Lesson Objectives

  • Solve quadratic equations by factoring.
  • Apply the Zero Product Property to find solutions to quadratic equations.
  • Analyze solutions and interpret them in real-world contexts.

Common Core Standards

  • HSA.SSE.3a: Factor a quadratic expression to reveal the zeros of the function it defines.
  • HSA.REI.4a: Solve quadratic equations in one variable.

Prerequisite Skills

  • Factoring polynomials.
  • Understanding of the Zero Product Property.
  • Basic multiplication skills.

Key Vocabulary

Additional Multimedia Resources

 

 

Warm Up Activities

Choose from one or more of these activities.

Activity 1: Factoring Quadratic Expressions Using Desmos

Students will practice factoring simpler polynomials using the Desmos graphing calculator.
Steps:

  1. Go to Desmos.com.
  2. Input quadratic function y = x2 + 5x + 6. 
  3. Have students identify the x-coordinates where the parabola intersects the x-axis.
  4. Then have students graph y = (x - 2)(x - 3).
  5. Have students discuss why the graphs are the same.
  6. Try other quadratic functions in standard form that also have integer x-intercepts.

 

Quadratics

 

Activity 2: Factor Tree and Prime Factorization

Students will factor different numbers using a factor tree to break them down into their prime factors. Review the following definitions before doing this activity:

Steps:

  1. Assign numbers such as 48, 60, 72, and 100.
  2. Have students start by creating factor pairs (e.g., 48 = 6• 8).
  3. Continue creating factor pairs each number until all branches end in prime numbers (e.g., 48 = (3•2)(2•2•2).
  4. Ask students to write the prime factorization of each number at the bottom of their template (e.g., 48 = 24 • 3<).
  5. Discuss how this skill connects to the process of factoring polynomials later in the lesson.

Activity 3: Factor It Out!

Objective: Help students practice rewriting variable expressions in factored form by identifying and factoring out common terms, including variables.

  • Begin by reviewing the concept of factoring out the greatest common factor (GCF).
  • Write a simple example on the board:
    • Example: 4x + 8
    • Ask: "What is the greatest common factor of 4x and 8?" (Answer: 4)
    • Rewrite: 4x + 8 = 4(x + 2).

Provide the following expressions on the board or as a handout. Students will rewrite each in factored form:

  1. 5x + 10
  2. 6x2 + 12x
  3. 3x + 9y
  4. 2x2 + 4x + 6
  5. x2 + 2x

Pair-and-Share

  • Have students work in pairs to solve these problems. Encourage them to compare answers and correct each other's work.
  • Remind students to identify both numerical and variable common factors when rewriting the expressions.

Quick Check 

  • Go over the answers as a class. Reinforce why the factored form is equivalent to the original expression.

Example Solutions

  • 5x + 10 = 5(x + 2)
  • 6x2 + 12x = 6x(x + 2)
  • 3x + 9y = 3(x + 3y)
  • 2x2 + 4x + 6 = 2(x^2 + 2x + 3)
  • x2 + 2x = x(x + 2)

Purpose: This activity ensures students can confidently identify and factor out GCFs from variable expressions. It serves as a bridge to more complex factoring tasks, such as factoring quadratic expressions later in the lesson.

 

 

Teach

Understanding the Zero Product Property

The Zero Product Property is a fundamental concept in solving quadratic equations. This property states that if the product of two or more factors equals zero, then at least one of the factors must be zero. Mathematically, 

if A · B = 0, then either A = 0 or B = 0

When applied to quadratics in factored form, this property becomes a powerful tool for finding solutions to quadratic solutions. For instance, consider the quadratic equation 

(x - 3)(x + 2) = 0

By the Zero Product Property, either 

x - 3 = 0 
or 
x + 2 = 0. 

Solving these equations gives the roots of the quadratic: x = 3 and x = -2.

In general, if a quadratic equation can be expressed as the product of two binomials, such as (x - a)(x - b) = 0, or as the product of a monomial and a binomial, such as x(x - c) = 0, the Zero Product Property provides a straightforward method for finding the roots. This approach not only simplifies the solution process but also highlights the connection between factoring and solving quadratics.

In this lesson, we will explore how to apply the Zero Product Property to quadratics in factored form and demonstrate its effectiveness in finding the roots of quadratic equations. 

Example 1

Solve x2 - 5x + 6 = 0

Factor x2 - 5x + 6 = 0

  1. Identify the quadratic form.

    The standard form of a quadratic equation is:

    ax2 + bx + c = 0

    In this case: a = 1, b = -5, and c = 6.

  2. Find two numbers whose product is 6 and whose sum is -5.

    The pair of numbers is -2 and -3 because:

-2 • (-3) = 6
-2 + (-3) = -5

  1. Write the factored form.

    Using the two numbers -2 and -3, the factored form is:

    x2 - 5x + 6 = (x - 2)(x - 3)

Solve the equation

Since the equation is set equal to zero, solve for x:

(x - 2)(x - 3) = 0

  • Set each factor equal to zero:

x - 2 = 0 or x - 3 = 0

  • Solve for x:

x = 2

x = 3

Summary: This example demonstrates the process of factoring a simple quadratic equation and finding its roots using the Zero Product Property.

Example 2

A ball is thrown upward, and its height is given by h(t) = -16t2 + 32t + 48. When does it hit the ground?

Solution:

  1. Write the equation: h(t) = -16t2 + 32t + 48.
  2. The ball hits the ground when h(t) = 0. So solve this equation:

-16t2 + 32t + 48 = 0.

  1. Divide through by -16: t2 - 2t - 3 = 0.
  2. To factor, find two numbers whose product is -2 and product is -3. The two numbers are -3 and 1:

-3 + 1 = -2
-3 • 1 = -3

  1. Factor the quadratic: (t - 3)(t + 1) = 0.
  2. Apply the Zero Product Property: t - 3 = 0 or t + 1 = 0.
  3. Solve for t: t = 3 (valid) and t = -1 (discarded).

Summary: This example applies factoring to a real-world scenario involving projectile motion. A negative root was rejected, since t must be a positive value.

 

Quadratics

 

Example 3

Solve 2x2 + 4x - 6 = 0

Solution:

  1. Write the equation: 2x2 + 4x - 6 = 0.
  2. Factor out the greatest common factor (GCF): 2(x2 + 2x - 3) = 0.
  3. Factor the quadratic: 2(x + 3)(x - 1) = 0.
  4. Apply the Zero Product Property: x + 3 = 0 or x - 1 = 0.
  5. Solve for x: x = -3 and x = 1.

Summary: This example illustrates factoring with a common factor and finding the solutions.

Example 4

A rectangular garden has an area of 30 square meters. Its length is x + 5 meters and its width is x meters. Find the dimensions of the garden.

 

Quadratics

 

  1. Equation: x(x + 5) = 30.
  2. Simplify: x2 + 5x - 30 = 0.
  3. Factor: (x + 10)(x - 3) = 0.
  4. Solutions: x = -10 (not valid), x = 3.
  5. Dimensions: 3 meters by 8 meters.

Summary: This example shows how to solve an area problem by solving a quadratic equation.

Additional Multimedia Resources

 

 

Review

Summarize key concepts:

  • Quadratic Equation: A quadratic function set equal to zero to find the roots of the equation.
  • Factoring Quadratics: Breaking down a quadratic expression into the product of two binomial expressions.
  • Applying the Zero Product Property: Using the principle that if the product of two factors is zero, at least one of the factors must be zero.
  • Interpreting Solutions in Context: Understanding the practical implications of the solutions.

Reinforce the lesson objectives by summarizing the factoring process and reviewing vocabulary definitions. Provide two additional examples:

Example 1: Projectile Motion Example

A ball is thrown upwards, and its height (in feet) after x seconds is given by the equation:

h(x) = -16x2 + 48x + 64

Objective: Determine when the ball hits the ground.

  1. Set h(x) = 0:

    The ball hits the ground when its height is zero:

    -16x2 + 48x + 64 = 0

  2. Simplify the equation:

    Divide both sides by -16:

    x2 - 3x - 4 = 0

  3. Factor the quadratic equation. Find two factors whose sum is -3 and product is -4:

    (x - 4)(x + 1) = 0

  4. Solve for x:

    x - 4 = 0 or x + 1 = 0

    x = 4 or x = -1

    Since time cannot be negative, x = 4.

Answer: The ball hits the ground after 4 seconds.

2. Area Calculation Example

A rectangular cloth has dimensions x and x + 10, and its total area is 24 square units.

Objective: Find the dimensions of the cloth.

  1. Write the equation for the area:

    The area of the rectangle is:

    x(x + 10) = 24

    Expand:

    x2 + 10x = 24

  2. Rearrange into standard quadratic form:

    x2 + 10x - 24 = 0

  3. Factor the quadratic equation:

    (x + 12)(x - 2) = 0

  4. Solve for x:

    x + 12 = 0 or x - 2 = 0

    x = -12 or x = 2

    Since dimensions cannot be negative, x = 2.

  5. Find the dimensions:

    If x = 2, then:

    Dimensions = 2 units by 12 units

Answer: The dimensions of the cloth are 2 units by 12 units.

Additional Multimedia Resources

 

 

Quiz

Directions: Solve each quadratic equation by factoring.

  1. x2 - 9 = 0

     
  2. x2 + 5x + 6 = 0

     
  3. 2x2 - 8x = 0

     
  4. x2 - 4x - 12 = 0

     
  5. 3x2 - 3x = 0

     
  6. x2 + 6x + 9 = 0

     
  7. x2 - 16 = 0

     
  8. x2 + x - 12 = 0

     
  9. 4x2 - 25 = 0

     
  10. 2x2 + 8x + 6 = 0

Answer Key:

  1. x = ±3
  2. x = -2, -3
  3. x = 0, 4
  4. x = 6, -2
  5. x = 0, 1
  6. x = -3
  7. x = ±4
  8. x = 3, -4
  9. x = ±5/2
  10. x = -1, -3