Grade 9 Math • Algebra & Functions

Intro Solving Quadratics — I can solve simple quadratic equations by factoring when solutions are integers at an introductory level.

Learn to crack open quadratic equations by finding two numbers that multiply and add just right.

Grade 9 Math emblem with a calculator, geometric shapes, and line graph
Inside this lesson 10 sections 5 questions 8 figures

SECTION 1

Historical Context & Motivation

Humans have been solving equations with squared unknowns for thousands of years. Ancient civilizations needed these calculations for surprisingly practical reasons — measuring land, designing buildings, and calculating the paths of projectiles. The idea of a quadratic equation — an equation where the variable is raised to the second power — appeared long before anyone used modern algebraic symbols. Over centuries, mathematicians developed increasingly elegant methods to solve them, and factoring became one of the most intuitive approaches for equations with integer solutions.

~2000 BCE • Babylonian Clay Tablets

Babylonian scribes solved problems equivalent to quadratic equations using geometric cut-and-paste methods on clay tablets, centuries before algebra existed as a formal discipline.

~300 BCE • Euclid's Geometric Algebra

The Greek mathematician Euclid recast quadratic problems as constructions with rectangles and squares in his masterwork, Elements, linking algebra to geometry.

~820 CE • Al-Khwarizmi's Algebra

Persian scholar al-Khwarizmi wrote the first systematic treatise on solving quadratic equations. The word "algebra" itself comes from the Arabic title of his book.

1600s • Modern Symbolic Notation

Mathematicians like Descartes and Viète introduced x² notation and formalized factoring as a standard technique, making it accessible in the form we use today.

So here's the central question this lesson tackles: when you see an equation like x² − 5x + 6 = 0, how do you find the values of x that make it true? Factoring gives you a powerful, fast method — especially when those solutions turn out to be nice whole numbers. Let's learn how.

SECTION 2

Core Principles & Definitions

Before you can factor a quadratic, you need to understand a few building blocks. A quadratic equation is any equation that can be written in the form ax² + bx + c = 0, where a, b, and c are constants and a ≠ 0. The highest power of the variable is 2, which is what makes it "quadratic" (from the Latin quadratus, meaning "square"). In this introductory lesson, we focus on equations where a = 1, so the leading coefficient is simply 1.

1

Standard Form

A quadratic equation is in standard form when written as x² + bx + c = 0. The equation must equal zero before you factor.
2

Factoring

Factoring means rewriting a polynomial as a product of simpler expressions. For quadratics, we write x² + bx + c as (x + p)(x + q).
3

Zero Product Property

The Zero Product Property states: if A × B = 0, then A = 0 or B = 0. This is the key that unlocks the solutions once you've factored.
4

Solutions (Roots)

The solutions (also called roots or zeros) are the x-values that make the equation true. A quadratic equation can have up to two solutions.
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KEY TAKEAWAY

Think of factoring like reverse engineering a recipe. If someone hands you a cake (the quadratic expression), factoring means figuring out the exact ingredients (the two binomials) that were multiplied together to bake it. Once you know the ingredients, the Zero Product Property tells you exactly which values of x make the whole thing equal zero.

SECTION 3

Visual Explanation — The Factor Pair Approach

The diagram below illustrates the core process of factoring a quadratic like x² + bx + c. Your job is to find two integers, p and q, that satisfy two conditions simultaneously: they must multiply to give c and add to give b. When both conditions are met, you've cracked the factorization.

Factoring x² + bx + c = (x + p)(x + q)x² + bx + cFind p and q such that…PRODUCT CONDITIONp × q = cSUM CONDITIONp + q = b(x + p)(x + q) = 0x + p = 0 → x = −px + q = 0 → x = −q✓ Two solutions found via the Zero Product Property
The flowchart shows the full factoring process: start with x² + bx + c, find integers p and q that satisfy both the product condition (p × q = c) and the sum condition (p + q = b), then apply the Zero Product Property to find the two solutions.

Notice how the entire method hinges on two simple questions: "What multiplies to c?" and "What adds to b?" If you can find a pair of integers that answers both questions, the rest is just algebra. In the next section, we'll formalize the math behind each step.

SECTION 4

Mathematical Framework

Let's pin down the algebra. When you multiply two binomials using the FOIL method (First, Outer, Inner, Last), you can see exactly why the sum-and-product trick works. The expansion (x + p)(x + q) = x² + qx + px + pq = x² + (p + q)x + pq shows that the coefficient of x is the sum p + q, and the constant term is the product pq.

FOIL EXPANSION

(x+p)(x+q)=x2+(p+q)x+pq(x + p)(x + q) = x² + (p + q)x + pq

Comparing with x² + bx + c: b = p + q and c = p × q

ZERO PRODUCT PROPERTY

IfA×B=0,thenA=0orB=0If A × B = 0, then A = 0 or B = 0

Once you factor x² + bx + c = (x + p)(x + q) = 0, set each factor equal to zero: x + p = 0 gives x = −p, and x + q = 0 gives x = −q.

STANDARD FORM REQUIREMENT

ax2+bx+c=0(witha=1forthislesson)ax² + bx + c = 0 (with a = 1 for this lesson)

The equation must equal zero before you factor. If it doesn't, rearrange all terms to one side first.

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Watch Out!

A common mistake is to factor and then forget to set each factor equal to zero. Factoring alone doesn't give you the solutions — you must apply the Zero Product Property to finish the problem. Also, always move all terms to one side so the equation equals zero before you begin factoring.

SECTION 5

Finding the Right Factor Pairs

The trickiest part of factoring is finding the right pair of integers. A systematic approach makes this much easier. Start by listing all factor pairs of c (the constant term), including negative pairs when c is positive and mixed-sign pairs when c is negative. Then check which pair adds up to b. The diagram below walks through this process for x² − 5x + 6 = 0.

Example: Factor x² − 5x + 6b = −5, c = 6 → Find p and q where p × q = 6 and p + q = −5Factor Pairs of 6pqp + q167✗235✗−1−6−7✗−2−3−5✓Resultx² − 5x + 6 = (x − 2)(x − 3)Zero Product Propertyx − 2 = 0 or x − 3 = 0x = 2x = 3Verify on the Parabola y = x² − 5x + 6xyx = 2x = 3
Left: a table of all factor pairs of c = 6, checking which pair sums to b = −5. The winning pair (−2, −3) is highlighted in green. Right: the factored form, the two solutions, and a parabola confirming the roots cross the x-axis at x = 2 and x = 3.

Here are a few sign-pattern tips. When c is positive, both p and q share the same sign — both positive if b is positive, both negative if b is negative. When c is negative, one of p and q is positive and the other is negative. Keeping these patterns in mind cuts your search time in half.

Sign of bSign of cSigns of p and qExample
Positive (+)Positive (+)Both positivex² + 7x + 12 → (x + 3)(x + 4)
Negative (−)Positive (+)Both negativex² − 7x + 12 → (x − 3)(x − 4)
Positive (+)Negative (−)Mixed: larger one positivex² + 2x − 8 → (x + 4)(x − 2)
Negative (−)Negative (−)Mixed: larger one negativex² − 2x − 8 → (x − 4)(x + 2)

Sign patterns guide for factoring x² + bx + c

SECTION 6

Worked Example — Step by Step

Let's solve x² + 3x − 10 = 0 by factoring, walking through each step carefully.

Solve x² + 3x − 10 = 0

  1. 1

    Step 1 — Confirm Standard Form

    The equation is already in the form x² + bx + c = 0. Here, b = 3 and c = −10. The leading coefficient is 1, so we can proceed with simple factoring.
    b = 3, c = −10
  2. 2

    Step 2 — Identify Sign Pattern

    Since c is negative (−10), the two numbers p and q must have opposite signs. Since b is positive (+3), the number with the larger absolute value must be positive.
    One positive, one negative (positive is larger)
  3. 3

    Step 3 — List Factor Pairs of −10

    The factor pairs of −10 with opposite signs are: (1, −10), (−1, 10), (2, −5), (−2, 5). Now check which pair sums to b = 3. Testing: 1 + (−10) = −9 ✗, (−1) + 10 = 9 ✗, 2 + (−5) = −3 ✗, (−2) + 5 = 3 ✓.
    p = −2, q = 5
  4. 4

    Step 4 — Write the Factored Form

    Replace the quadratic with two binomials using p and q: x² + 3x − 10 = (x − 2)(x + 5).
    (x − 2)(x + 5) = 0
  5. 5

    Step 5 — Apply the Zero Product Property

    Set each factor equal to zero. First factor: x − 2 = 0, so x = 2. Second factor: x + 5 = 0, so x = −5.
    x = 2 or x = −5
  6. 6

    Step 6 — Verify Both Solutions

    Check x = 2: (2)² + 3(2) − 10 = 4 + 6 − 10 = 0 ✓. Check x = −5: (−5)² + 3(−5) − 10 = 25 − 15 − 10 = 0 ✓. Both solutions work.
    Confirmed: x = 2 and x = −5

SECTION 7

Tips, Strengths & Common Pitfalls

Factoring is a fast and elegant method, but it has its sweet spot. Understanding where it shines and where it struggles will help you choose the right tool for each problem.

StrengthsLimitations
Very fast once you spot the factor pair — often solvable in under a minute.Only works cleanly when the solutions are integers (or simple fractions).
No formula to memorize — just multiply/add logic.Doesn't help when c has many factor pairs, making trial-and-error slow.
Builds strong number sense and understanding of polynomial structure.Equations must be rearranged to standard form (= 0) first, which students sometimes forget.
Naturally reveals the connection between roots and factors of a polynomial.When a ≠ 1, factoring gets significantly harder and requires additional techniques.

When factoring works well vs. when you may need another method

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AVOID THESE COMMON MISTAKES

Mistake 1: Forgetting to set the equation equal to zero before factoring. If your equation is x² + 3x = 10, you must first rewrite it as x² + 3x − 10 = 0. Mistake 2: Finding the factors but stopping there — you must set each factor equal to zero. Mistake 3: Sign errors — always double-check by plugging solutions back into the original equation.

SECTION 8

Connection to Advanced Methods

Factoring is your first tool for solving quadratics, but it won't work for every equation. As you progress in algebra, you'll encounter quadratics with non-integer or irrational solutions — like x² − 3x + 1 = 0 — where factoring over the integers is impossible. That's when you'll learn more powerful methods.

MethodWhen to Use ItKey Idea
Factoring (this lesson)Solutions are integers (or simple fractions); a = 1Find p and q that multiply to c and add to b
Completing the SquareAny quadratic; useful for deriving vertex formRewrite as (x + h)² = k, then take the square root
Quadratic FormulaAny quadratic — works every time, even for irrational or complex rootsx = (−b ± √(b² − 4ac)) / 2a
GraphingEstimating solutions visually or with technologyFind x-intercepts of the parabola y = ax² + bx + c

Comparison of quadratic-solving methods

Even after you learn these advanced techniques, factoring remains valuable. It builds the intuition you need for completing the square, and the Zero Product Property shows up again in higher-level math when you solve polynomial equations of degree 3 and beyond. Mastering factoring now sets a strong foundation for everything that follows.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain in your own words why the equation must be set equal to zero before you can use factoring to solve it. What property depends on this condition?
Show answer
The Zero Product Property states that if A × B = 0, then at least one of A or B must be zero. This property only works when the product equals zero — if the product equaled some other number, like 6, you couldn't conclude anything specific about A or B (for example, A could be 2 and B could be 3, or A could be 1 and B could be 6). So we need the equation in the form (factor)(factor) = 0 for the property to apply, which requires moving all terms to one side first.

Question 2: BASIC CALCULATION

Solve by factoring: x² + 9x + 14 = 0.
Show answer
We need two numbers that multiply to 14 and add to 9. The factor pairs of 14 are (1, 14) and (2, 7). Since 2 + 7 = 9, the factorization is (x + 2)(x + 7) = 0. Setting each factor equal to zero: x + 2 = 0 gives x = −2, and x + 7 = 0 gives x = −7. Verification: (−2)² + 9(−2) + 14 = 4 − 18 + 14 = 0 ✓ and (−7)² + 9(−7) + 14 = 49 − 63 + 14 = 0 ✓. The solutions are x = −2 and x = −7.

Question 3: INTERMEDIATE

Solve by factoring: x² − 4x − 21 = 0.
Show answer
Here b = −4 and c = −21. Since c is negative, the two numbers have opposite signs, and since b is negative, the number with the larger absolute value is negative. Factor pairs of −21 with opposite signs: (1, −21), (−1, 21), (3, −7), (−3, 7). Testing sums: 3 + (−7) = −4 ✓. So the factorization is (x + 3)(x − 7) = 0. Setting each factor to zero: x + 3 = 0 gives x = −3, and x − 7 = 0 gives x = 7. Check: (−3)² − 4(−3) − 21 = 9 + 12 − 21 = 0 ✓ and (7)² − 4(7) − 21 = 49 − 28 − 21 = 0 ✓. The solutions are x = −3 and x = 7.

Question 4: APPLIED

A rectangular garden has a length that is 3 meters more than its width. The area of the garden is 40 square meters. Write a quadratic equation and solve it by factoring to find the garden's dimensions.
Show answer
Let w represent the width. Then the length is w + 3, and the area equation is w(w + 3) = 40. Expanding: w² + 3w = 40. Moving to standard form: w² + 3w − 40 = 0. We need two numbers that multiply to −40 and add to 3. Testing factor pairs: (8, −5) gives 8 + (−5) = 3 ✓. The factorization is (w + 8)(w − 5) = 0, yielding w = −8 or w = 5. Since width cannot be negative, w = 5 meters. The length is 5 + 3 = 8 meters. Check: 5 × 8 = 40 ✓. The garden is 5 m by 8 m.

Question 5: CRITICAL THINKING

A student claims that the equation x² − 6x + 9 = 0 has only one solution. Is this correct? Factor the equation, explain what happens, and describe what this means graphically for the parabola y = x² − 6x + 9.
Show answer
The student is essentially correct. We need two numbers that multiply to 9 and add to −6. The pair (−3, −3) works: (−3) × (−3) = 9 and (−3) + (−3) = −6. So x² − 6x + 9 = (x − 3)(x − 3) = (x − 3)² = 0. Setting x − 3 = 0 gives x = 3 as the only solution. This is called a repeated root or a double root. Graphically, the parabola y = x² − 6x + 9 touches the x-axis at exactly one point — the vertex sits right on the x-axis at (3, 0). Instead of crossing the axis at two distinct points, the parabola just "bounces" off it. This happens whenever the quadratic is a perfect square trinomial.

SUMMARY

Lesson Summary

Summary

To solve a quadratic equation of the form x² + bx + c = 0 by factoring, find two integers p and q that satisfy the product condition (p × q = c) and the sum condition (p + q = b). Rewrite the quadratic as (x + p)(x + q) = 0, then apply the Zero Product Property to set each factor equal to zero, giving solutions x = −p and x = −q.

Use the sign pattern of b and c to narrow your search: when c is positive, both numbers share a sign; when c is negative, the numbers have opposite signs. Always confirm your equation is in standard form (equal to zero) before factoring, and always verify by substituting your solutions back into the original equation. Factoring is fastest when solutions are integers, and it builds the foundation for advanced methods like completing the square and the quadratic formula.

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