SAT Math Lesson Plan 15: Quadratic Equations and Factoring

Lesson Summary

Lesson 15 in this SAT Math Prep series focuses on quadratic equations and factoring—concepts that appear in approximately 15–20% of SAT Math questions. Students will learn how to factor quadratic expressions of the form \( ax^2 + bx + c \), solve equations using the zero-product property, and recognize when factoring is the most efficient method. Real-world applications from physics and engineering will demonstrate how quadratic models are used in practical scenarios. This 45-minute lesson includes multiple worked examples, practice problems, a quiz, and an answer key. The content is aligned with Common Core standards and reinforces algebraic fluency needed for the SAT.

Lesson Objectives

Factor quadratic expressions in the form \( ax^2 + bx + c \).
Solve quadratic equations by factoring and applying the zero-product property.
Recognize when factoring is the most efficient method for solving a quadratic.
Apply quadratic equations to model and solve real-world problems.

Common Core Standards

CCSS.MATH.CONTENT.HSA.SSE.B.3.A – Factor a quadratic expression to reveal the zeros of the function it defines.
CCSS.MATH.CONTENT.HSA.REI.B.4.B – Solve quadratic equations by inspection, factoring, completing the square, and the quadratic formula.

Prerequisite Skills

Basic algebraic operations and simplification
Knowledge of exponents and the distributive property
Experience solving linear and simple quadratic equations

Key Vocabulary

Quadratic Equation: An equation in the form \( ax^2 + bx + c = 0 \).
Factoring: Expressing a quadratic as the product of two binomials.
Zero-Product Property: If \( ab = 0 \), then \( a = 0 \) or \( b = 0 \).
Perfect Square Trinomials: Expressions like \( x^2 + 2ax + a^2 = (x + a)^2 \).
Difference of Squares: A binomial of the form \( a^2 - b^2 = (a - b)(a + b) \).

Warm Up

This warm-up includes two activities to review factoring of basic quadratics and introduces the zero-product property.

Activity 1: Factoring Review

Instructions: Review how to factor quadratic expressions of the form \( x^2 + bx + c \). These expressions can often be written as the product of two binomials: \[ x^2 + bx + c = (x + r_1)(x + r_2) \] Where:

\( r_1 + r_2 = b \)
\( r_1 \cdot r_2 = c \)

Practice: Factor the following quadratics by identifying values of \( r_1 \) and \( r_2 \) such that their product is \( c \) and their sum is \( b \).

\( x^2 + 5x + 6 \)
\( x^2 - 3x + 2 \)
\( x^2 + 7x + 10 \)

Solutions:

\( x^2 + 5x + 6 = (x + 2)(x + 3) \)
\( x^2 - 3x + 2 = (x - 1)(x - 2) \)
\( x^2 + 7x + 10 = (x + 2)(x + 5) \)

Self-Study Tip: Practice spotting factorable quadratics quickly. This will help you solve SAT problems more efficiently—especially when factoring is the simplest method.

Activity 2: Equation Solving with the Zero-Product Property

Problem: Solve the equation \( x^2 + 5x + 6 = 0 \).

Step 1:

Factor the expression: \[ x^2 + 5x + 6 = (x + 2)(x + 3) \]

Step 2:

Apply the zero-product property: \[ (x + 2)(x + 3) = 0 \Rightarrow x = -2 \text{ or } x = -3 \]

Answer: \( x = -2 \), \( x = -3 \)

Teach

Factoring is one of the most efficient methods for solving quadratic equations—especially when the equation is already in the form \( x^2 + bx + c = 0 \). In the Warm-Up, you practiced identifying two numbers whose product is \( c \) and whose sum is \( b \), allowing you to rewrite the quadratic as a product of two binomials. This technique is a valuable time-saver on the SAT, where many quadratics are designed to factor cleanly.

In this section, we’ll build on that foundational skill and explore several key types of quadratics that also factor easily. These include:

Trinomials with leading coefficients other than 1
Perfect square trinomials
Differences of squares
Quadratics embedded in real-world applications

Each example will show you how to spot a pattern, select the appropriate factoring strategy, and solve for the variable(s). Understanding how to quickly recognize a factored form and apply the zero-product property will help you tackle a wide range of SAT Math problems efficiently.

Example 1: Factoring a Simple Trinomial

Problem: Solve the equation \( x^2 + 7x + 12 = 0 \).

Step 1: Factor the Trinomial

We're looking for two numbers:

Whose product is 12 (the constant term)
Whose sum is 7 (the coefficient of \( x \))

Try the factor pairs of 12:

\( 1 \times 12 \): \( 1 + 12 = 13 \)
\( 2 \times 6 \): \( 2 + 6 = 8 \)
\( 3 \times 4 \): \( 3 + 4 = 7 \) ✅

So the correct factorization is: \[ x^2 + 7x + 12 = (x + 3)(x + 4) \]

Step 2: Apply the Zero-Product Property

\[ (x + 3)(x + 4) = 0 \Rightarrow x + 3 = 0 \quad \text{or} \quad x + 4 = 0 \]

Step 3: Solve Each Equation

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

Answer: \( x = -3 \) or \( x = -4 \)

This is a classic SAT-style factoring problem. When the leading coefficient is 1, use the product-sum method to factor efficiently.

Example 2: Factoring with a Leading Coefficient Greater Than 1

Problem: Solve \( 2x^2 + 7x + 3 = 0 \).

Step 1: Multiply the Leading Coefficient and Constant

\[ 2 \times 3 = 6 \] We now look for two numbers:

Whose product is 6
Whose sum is 7 (the coefficient of the middle term)

Factor pairs of 6:

\( 1 \times 6 = 6 \), and \( 1 + 6 = 7 \) ✅
\( 2 \times 3 = 6 \), and \( 2 + 3 = 5 \)

So we’ll use 1 and 6.

Step 2: Rewrite the Middle Term Using the Two Numbers

We now break up the middle term \( 7x \) into \( 1x + 6x \), since: \[ 1x + 6x = 7x \quad \text{and} \quad 1x \cdot 6x = 6x^2 \] This gives us: \[ 2x^2 + x + 6x + 3 = 0 \]

Step 3: Group the Terms in Pairs

Group the first two terms and the last two terms so we can factor each group: \[ (2x^2 + x) + (6x + 3) \] Now factor each group:

From \( 2x^2 + x \), factor out \( x \): \( x(2x + 1) \)
From \( 6x + 3 \), factor out 3: \( 3(2x + 1) \)

So now we have: \[ x(2x + 1) + 3(2x + 1) \]

Step 4: Factor the Common Binomial

Since both terms contain \( (2x + 1) \), factor that out: \[ (x + 3)(2x + 1) = 0 \]

Step 5: Solve Each Equation

Use the zero-product property: \[ x + 3 = 0 \Rightarrow x = -3 \quad ; \quad 2x + 1 = 0 \Rightarrow x = -\frac{1}{2} \]

Answer: \( x = -3 \) or \( x = -\frac{1}{2} \)

Factoring by grouping is a key strategy when the leading coefficient is not 1. Break apart the middle term using two numbers that meet the product-sum requirement, then factor step-by-step.

Teacher Note: On the SAT, factoring is almost always the fastest method for solving a quadratic equation—especially when integer solutions are expected. Encourage students to look for opportunities to factor before considering the quadratic formula or completing the square. If the expression doesn’t factor easily, then switch methods.

Example 3: Difference of Squares

Problem: Solve \( x^2 - 49 = 0 \).

Step 1:

Recognize this as a difference of squares: \[ x^2 - 49 = (x - 7)(x + 7) \]

Step 2:

Apply the zero-product property: \[ x - 7 = 0 \quad \text{or} \quad x + 7 = 0 \]

Step 3:

\[ x = 7 \quad \text{or} \quad x = -7 \]

Answer: \( x = 7 \) or \( x = -7 \)

Always look for recognizable patterns. The difference of squares is a shortcut that can save time on the SAT.

Extension: The difference of squares pattern also applies to higher-degree expressions. For example:

\[ 64x^4 - 16 = (8x^2)^2 - 4^2 = (8x^2 - 4)(8x^2 + 4) \]

This can be factored further:

\[ = 4(2x^2 - 1) \cdot 4(2x^2 + 1) = 16(2x^2 - 1)(2x^2 + 1) \]

While this level of factoring is rarely tested on the SAT, it’s a good way to reinforce how the difference of squares pattern works—even with fourth powers.

Example 4: Perfect Square Trinomial

Problem: Solve \( x^2 - 6x + 9 = 0 \).

Step 1:

Recognize the trinomial as a perfect square: \[ x^2 - 6x + 9 = (x - 3)^2 \]

Step 2:

Set the factored expression equal to zero: \[ (x - 3)^2 = 0 \]

Step 3:

Take the square root of both sides: \[ x - 3 = 0 \Rightarrow x = 3 \]

Answer: \( x = 3 \)

When a quadratic factors into a perfect square binomial, it has only one solution (a repeated root). The SAT may test this with both algebraic expressions and graphs.

Example 5: Real-World Application

Problem: The length of a rectangular garden is 3 feet more than its width. If the area is 40 square feet, what are the dimensions of the garden?

Step 1:

Let the width be \( x \). Then the length is \( x + 3 \). Area = length × width: \[ x(x + 3) = 40 \]

Step 2:

Distribute: \[ x^2 + 3x = 40 \]

Step 3:

Rearrange the equation: \[ x^2 + 3x - 40 = 0 \]

Step 4:

Factor: \[ (x + 8)(x - 5) = 0 \]

Step 5:

Solve: \[ x = -8 \quad \text{(not valid)} \quad \text{or} \quad x = 5 \]

Step 6:

Width = 5 ft; Length = 8 ft

Answer: The dimensions are 5 feet by 8 feet.

The SAT often embeds quadratic equations in geometry and measurement contexts. Be prepared to translate verbal information into algebraic equations.

Review

This lesson introduced key strategies for solving quadratic equations by factoring—an essential algebra skill that appears frequently on the SAT. You practiced factoring trinomials, identifying special patterns like perfect squares and differences of squares, and solving real-world problems using quadratic models. Factoring is often the fastest method for solving quadratic equations, especially when the equation is factorable with integer roots.

Here’s a summary of what you learned:

How to factor quadratics of the form \( ax^2 + bx + c \)
How to recognize and use patterns such as perfect square trinomials and differences of squares
How to apply the zero-product property to solve equations
How to model and solve real-world problems using quadratic equations

You also learned how to recognize and factor perfect square trinomials. These are quadratics in the form: \[ x^2 \pm 2ax + a^2 = (x \pm a)^2 \] This structure shows up frequently on the SAT and often results in a single solution, because the two factors are identical. When graphed, perfect square trinomials correspond to parabolas that touch the x-axis at one point—called a repeated root. While this lesson focused on algebraic factoring, it’s helpful to remember that the factored form also reveals important information about the graph.

Example 1: Basic Quadratic

Problem: Solve \( x^2 - 2x - 15 = 0 \).

Step 1:

Factor the trinomial: \[ x^2 - 2x - 15 = (x - 5)(x + 3) \]

Step 2:

Use the zero-product property: \[ x - 5 = 0 \Rightarrow x = 5 \quad ; \quad x + 3 = 0 \Rightarrow x = -3 \]

Answer: \( x = 5 \) or \( x = -3 \)

Example 2: Leading Coefficient Not Equal to 1

Problem: Solve \( 3x^2 - x - 2 = 0 \).

Step 1:

Multiply: \( 3 \times -2 = -6 \)
Find two numbers that multiply to -6 and add to -1: -3 and 2

Step 2:

Rewrite the middle term: \[ 3x^2 - 3x + 2x - 2 = 0 \]

Step 3:

Factor by grouping: \[ 3x(x - 1) + 2(x - 1) = (3x + 2)(x - 1) \]

Step 4:

\[ 3x + 2 = 0 \Rightarrow x = -\frac{2}{3} \quad ; \quad x - 1 = 0 \Rightarrow x = 1 \]

Answer: \( x = -\frac{2}{3} \) or \( x = 1 \)

Example 3: Geometry Application

Problem: A square patio has an area of 64 square feet. A border of uniform width is added around the patio, increasing the total area to 100 square feet. What is the width of the border?

Step 1:

Let the border width be \( x \). Then the total side length is \( 8 + 2x \). Set up the equation: \[ (8 + 2x)^2 = 100 \]

Step 2:

\[ 64 + 32x + 4x^2 = 100 \Rightarrow 4x^2 + 32x - 36 = 0 \]

Step 3:

Divide all terms by 4: \[ x^2 + 8x - 9 = 0 \]

Step 4:

Factor: \[ (x + 9)(x - 1) = 0 \]

Step 5:

\[ x = -9 \text{ (not valid)} \quad ; \quad x = 1 \]

Answer: The width of the border is 1 foot.

Multimedia Resources

For additional videos, tutorials, and interactives to reinforce the concepts in this lesson, explore this curated set of resources from Media4Math.

Click here to access the Lesson 15 Multimedia Resources.

Quiz

Directions: Choose the best answer for each question. Show your work on a separate sheet if needed.

Solve: \( x^2 + 6x + 8 = 0 \)
a) -4, -2
b) -3, -2
c) -4, 2
d) -2, 4
Solve: \( x^2 - 16 = 0 \)
a) -8, 8
b) -4, 4
c) -2, 2
d) -6, 6
Solve: \( 2x^2 + 7x + 3 = 0 \)
a) -3, -2
b) -1, -3
c) -3, -1/2
d) -1/2, -3
Which of the following is a factorization of \( x^2 + 4x + 4 \)?
a) \( (x + 2)^2 \)
b) \( (x + 4)(x + 1) \)
c) \( (x + 1)^2 \)
d) \( (x + 2)(x - 2) \)
Factor: \( 3x^2 - 2x - 1 \)
a) \( (3x + 1)(x - 1) \)
b) \( (3x - 1)(x + 1) \)
c) \( (3x + 1)(x + 1) \)
d) \( (x - 1)(3x - 1) \)
Solve: \( x^2 + x = 0 \)
a) 0, 1
b) 0, -1
c) 1, -1
d) 0, 2
Which value of \( x \) satisfies \( (x - 3)^2 = 0 \)?
a) -3
b) 3
c) 0
d) 9
Solve: \( 4x^2 - 25 = 0 \)
a) -4, 5
b) -5, 5
c) -2, 5
d) -5/2, 5/2
Which of the following is equivalent to \( x^2 - 2x + 1 \)?
a) \( (x - 2)^2 \)
b) \( (x + 1)^2 \)
c) \( (x - 1)^2 \)
d) \( (x - 3)^2 \)
What are the solutions to \( 5x^2 + 10x = 0 \)?
a) -2, 0
b) -5, 0
c) 0, 2
d) 0, -1
A rectangular garden has a length that is 5 feet longer than its width. If the area of the garden is 36 square feet, what is the width of the garden?
a) 4 ft
b) 5 ft
c) 6 ft
d) 9 ft
A ball is thrown upward from the ground and its height \( h \) in feet after \( t \) seconds is modeled by the equation \( h = -16t^2 + 32t \). When does the ball return to the ground?
a) 0 seconds
b) 1 second
c) 2 seconds
d) 4 seconds

Answer Key

Answer: a) -4, -2
\[ x^2 + 6x + 8 = (x + 4)(x + 2) \Rightarrow x = -4, -2 \]
Answer: b) -4, 4
\[ x^2 - 16 = (x - 4)(x + 4) \Rightarrow x = \pm 4 \]
Answer: d) -1/2, -3
\[ 2x^2 + 7x + 3 = (2x + 1)(x + 3) \Rightarrow x = -\frac{1}{2}, -3 \]
Answer: a) \( (x + 2)^2 \)
\[ x^2 + 4x + 4 = (x + 2)^2 \]
Answer: b) \( (3x - 1)(x + 1) \)
\[ 3x^2 - 2x - 1 = (3x - 1)(x + 1) \]
Answer: b) 0, -1
\[ x^2 + x = x(x + 1) \Rightarrow x = 0, -1 \]
Answer: b) 3
\[ (x - 3)^2 = 0 \Rightarrow x - 3 = 0 \Rightarrow x = 3 \]
Answer: d) -5/2, 5/2
\[ 4x^2 - 25 = (2x - 5)(2x + 5) \Rightarrow x = \pm \frac{5}{2} \]
Answer: c) \( (x - 1)^2 \)
\[ x^2 - 2x + 1 = (x - 1)^2 \]
Answer: a) -2, 0
\[ 5x^2 + 10x = 5x(x + 2) \Rightarrow x = 0, -2 \]
Answer: a) 4 ft
Let width = \( x \), then length = \( x + 5 \). Area: \[ x(x + 5) = 36 \Rightarrow x^2 + 5x - 36 = 0 \] Factor: \[ (x + 9)(x - 4) = 0 \Rightarrow x = -9 \text{ (reject), } x = 4 \]
Answer: c) 2 seconds
Set height equal to 0 (ground level): \[ -16t^2 + 32t = 0 \Rightarrow -16t(t - 2) = 0 \Rightarrow t = 0 \text{ or } t = 2 \] The ball hits the ground again after **2 seconds**.