HighSchoolQ

➕ Algebra 1 · Quadratics

Solving Quadratics by Factoring

Solve a quadratic equation by setting one side to zero, factoring, and setting each factor to zero. Includes double roots, x = 0, graphs and area problems.

Grades 9, 10⏱ 17 min

By the end you can…

  • Explain why the zero-product property works, and why an equation must equal 0 on one side before you use it.
  • Solve a quadratic equation by writing it as ax2 + bx + c = 0, factoring completely, setting each factor equal to 0, solving, and checking.
  • Handle special cases: a double root, a common factor that gives the solution x = 0, a difference of squares, and an equation that does not factor with integers.
  • Connect the solutions to the x-intercepts of a parabola, and solve area problems, rejecting a solution that cannot fit the situation.

Lesson (school style)

A quadratic equation is an equation that can be written as ax2 + bx + c = 0 with a ≠ 0. Its solutions are the values of x that make it true. You already know how to factor expressions. This lesson uses factoring to solve: if a product equals 0, one of its factors must equal 0, so one hard equation turns into two easy ones. The method has one strict requirement: the product must equal 0, so get 0 on one side first.

Quadratic equations and their solutions

A solution (also called a root) of an equation is a number that makes it true when you substitute it. For x2 − 5x + 6 = 0: x = 2 is a solution because 4 − 10 + 6 = 0, and x = 3 is a solution because 9 − 15 + 6 = 0. But x = 4 is not, because 16 − 20 + 6 = 2.

A quadratic equation has at most two real solutions. (A real number is any number on the number line. In this course every solution is a real number.) Sometimes it has two, sometimes one (a double root, below), and sometimes none. For example, x2 + 1 = 0 has no real solution, because x2 is never negative, so x2 + 1 is always at least 1.

The zero-product property

Zero-product property: if A × B = 0, then A = 0 or B = 0 (or both). Here A and B can be numbers or whole expressions, like x − 2.

Why does it work? Suppose A × B = 0.

  • If A = 0, then one factor is 0, and we are done.
  • If A ≠ 0, we are allowed to divide both sides by A (dividing by a number that is not zero is fine). That gives B = 0 ÷ A = 0.

Either way, at least one factor is 0. In words: you cannot multiply two nonzero numbers and get 0.

The same idea works for three or more factors: if A × B × C = 0, at least one of them is 0.

Why one side must be 0

Zero is the only number with this power. A product equal to 12 could be 3 × 4, or 2 × 6, or (1/2) × 24, so it tells you nothing about what one factor equals on its own.

Look at (x + 1)(x + 4) = 10. A tempting move is to set each factor equal to 10. That gives x = 9 or x = 6, but both are wrong. At x = 9 the product is 10 × 13 = 130, and at x = 6 it is 7 × 10 = 70, not 10.

The right way is to multiply out and move the 10 first: x2 + 5x + 4 = 10, so x2 + 5x − 6 = 0. Now factor: (x + 6)(x − 1) = 0. So x = −6 or x = 1. Check: (−6 + 1)(−6 + 4) = (−5)(−2) = 10 and (1 + 1)(1 + 4) = 2 × 5 = 10. Both work.

The five steps

  1. Zero on one side. Move every term to one side, so the equation looks like ax2 + bx + c = 0. If a is negative, you can multiply both sides by −1 (allowed, since −1 is not 0) to make the x2 term positive. That changes the sign of every term.
  2. Factor completely. Take out the greatest common factor first. Then factor what is left (a trinomial, a difference of squares, and so on). Completely factored means that no factor can be factored further using integers.
  3. Set each factor that contains x equal to 0. This is the zero-product property. A factor that is just a number, like 3, can never be 0, so it gives no equation.
  4. Solve each equation. Each one is a one- or two-step equation.
  5. Check each solution in the original equation. If you factored and solved correctly, every solution works. The check catches slips, such as a sign error in step 1 or a wrong factor pair.

Remember the sign flip: the factor x + 4 gives x = −4, and the factor x − 3 gives x = 3. (For a factor like 2x + 1, you also divide: 2x = −1, so x = −1/2.)

Special case: a double root

Sometimes the same factor appears twice. The trinomial x2 + 8x + 16 factors as (x + 4)(x + 4) = (x + 4)2. Setting each factor to 0 gives x + 4 = 0 both times, so the only solution is x = −4. This is called a double root (or repeated solution). It is listed once: the equation has one solution.

A double root shows up when the trinomial is a perfect square, like x2 + 8x + 16 or x2 − 10x + 25. On the graph of y = x2 + 8x + 16, the parabola touches the x-axis at one point, its vertex, and does not cross it.

A common factor, and the solution x = 0

Always look for a greatest common factor first. In 3x2 + 21x = 0 the GCF is 3x, so the equation becomes 3x(x + 7) = 0. There are three factors: 3, x and x + 7.

  • The factor 3 is never 0, so it gives no solution.
  • The factor x gives x = 0.
  • The factor x + 7 gives x = −7.

When x is a common factor, x = 0 is a solution. It is a perfectly good solution. Zero is a number like any other, and it makes the equation true: 3(0)2 + 21(0) = 0.

Do not divide both sides by x

Take x2 = 6x. It is tempting to divide both sides by x and get x = 6. But dividing by x is only allowed when x is not zero (assume any variable you divide by is not zero). Here x might be zero, and it is: 02 = 6(0) is true. Dividing by x quietly throws that solution away.

Factoring does not divide by x, so it keeps both solutions: x2 − 6x = 0, so x(x − 6) = 0, so x = 0 or x = 6.

Dividing both sides by a number that is not zero, such as 3 or −16, is always safe. The danger is dividing by an expression that could be 0.

Differences of squares and perfect squares

The patterns from earlier lessons make factoring fast. A difference of squares factors as a2 − b2 = (a + b)(a − b). So x2 − 49 = 0 becomes (x + 7)(x − 7) = 0, and the solutions are x = −7 and x = 7. Both signs appear: the equation x2 = 49 is true for x = 7 and for x = −7.

A perfect square trinomial, a2 ± 2ab + b2 = (a ± b)2, gives a double root when you set it equal to 0.

When the equation does not factor

Not every quadratic can be factored with integers. To factor x2 + 2x − 4, you would need two integers with product −4 and sum 2. The pairs with product −4 are 1 and −4, −1 and 4, and 2 and −2, with sums −3, 3 and 0. None adds up to 2. A trinomial like this is prime (it cannot be factored using integers), so this method stops.

That does not automatically mean the equation has no solution. For x2 + 2x − 4 = 0, the graph of y = x2 + 2x − 4 still crosses the x-axis, so solutions exist, but they are not whole numbers (see worked example 9). Whether an equation that does not factor has real solutions depends on the equation: x2 + 1 = 0 has none, for instance. Later lessons (completing the square, the quadratic formula and the discriminant) find the solutions exactly and tell you how many there are. So when you cannot find a factor pair, switch methods. Do not conclude that there is no answer.

Solutions, x-intercepts and the graph

The graph of y = ax2 + bx + c meets the x-axis where y = 0. Substituting y = 0 gives the equation ax2 + bx + c = 0. So the solutions of the equation are exactly the x-intercepts of the parabola (the x-values where it meets the x-axis, at the points (p, 0)). Another name for them is the zeros of the function.

  • Two different solutions: the parabola crosses the x-axis twice, and its axis of symmetry is halfway between them.
  • A double root: the parabola touches the x-axis once, at its vertex.
  • No real solution: the parabola never reaches the x-axis.

This link works only for the equation set equal to 0. The equation x2 − 4x + 3 = 8 asks where the parabola is 8 units high, which is not the same as asking where it meets the x-axis.

Area word problems

Word problems often lead to a quadratic equation, because area multiplies two lengths.

  1. Define the variable in words: "let w be the width in inches."
  2. Write the other measurements using the variable.
  3. Write an equation from the situation, such as length × width = area.
  4. Move everything to one side, factor, and solve.
  5. Check each solution against the story. The algebra may give a solution that the story cannot use. A length cannot be negative, so a negative solution for a width is rejected, and you should say why.
  6. Answer in a sentence with units.

A rejected solution still solves the equation. It just does not describe a real rectangle.

Key words

Quadratic equation
An equation that can be written as ax2 + bx + c = 0, where a ≠ 0.
Zero-product property
If A × B = 0, then A = 0 or B = 0 (or both). It works only when the product equals 0.
Solution (root)
A number that makes the equation true when you substitute it. For x2 − 5x + 6 = 0, the solutions are 2 and 3.
Double root
A solution that comes from a factor appearing twice, such as x = −4 from (x + 4)2 = 0. It is listed once.
Completely factored
Written as a product of any GCF and prime factors, so that no factor can be factored further using integers.
Prime (cannot be factored using integers)
A polynomial that cannot be written as a product of simpler polynomials with integer coefficients, other than by taking out 1 or −1. For example, x2 + x + 1 is prime. A polynomial with a GCF greater than 1, such as 2x2 + 4x + 6, is not prime: take out the GCF and test what is left.
x-intercept (zero)
An x-value where a graph meets the x-axis, so y = 0 there. If it is p, the graph passes through (p, 0). The solutions of ax2 + bx + c = 0 are the x-intercepts of y = ax2 + bx + c.

Worked examples

Example 1. Solve x2 + 2x − 24 = 0.

  1. The equation already equals 0. It is in standard form with a = 1, b = 2, c = −24.
  2. Factor: find two integers with product −24 and sum 2. The pair 6 and −4 works, because 6 × (−4) = −24 and 6 + (−4) = 2. So x2 + 2x − 24 = (x + 6)(x − 4).
  3. Zero-product property: (x + 6)(x − 4) = 0 means x + 6 = 0 or x − 4 = 0.
  4. Solve each: x = −6 or x = 4.
  5. Check x = −6: (−6)2 + 2(−6) − 24 = 36 − 12 − 24 = 0. Check x = 4: 16 + 8 − 24 = 0. Both are true.

Answer: Answer: x = −6 or x = 4

Example 2. Solve x2 = 2x + 15.

  1. The equation does not equal 0, so move everything to the left side. Subtract 2x and subtract 15 from both sides: x2 − 2x − 15 = 0.
  2. Factor: product −15 and sum −2. The pair 3 and −5 works. So (x + 3)(x − 5) = 0.
  3. Set each factor to 0: x + 3 = 0 or x − 5 = 0. So x = −3 or x = 5.
  4. Check in the original equation. For x = −3: left side (−3)2 = 9, right side 2(−3) + 15 = 9. For x = 5: left side 25, right side 2(5) + 15 = 25.

Answer: Answer: x = −3 or x = 5

Example 3. Solve −x2 + 2x + 8 = 0.

  1. The leading coefficient is negative. Multiply both sides by −1 (allowed, because −1 is not 0). Every term changes sign: x2 − 2x − 8 = 0.
  2. Factor: product −8 and sum −2. The pair 2 and −4 works. So (x + 2)(x − 4) = 0.
  3. Set each factor to 0: x = −2 or x = 4.
  4. Check in the original equation. For x = −2: −(−2)2 + 2(−2) + 8 = −4 − 4 + 8 = 0. For x = 4: −(4)2 + 2(4) + 8 = −16 + 8 + 8 = 0.

Answer: Answer: x = −2 or x = 4

Example 4. Solve 2x2 − 5x − 3 = 0.

  1. The equation equals 0, and there is no common factor. Here a = 2, so use the ac method from the factoring trinomials lesson (or trial and check): ac = 2 × (−3) = −6, and we need two integers with product −6 and sum −5. The pair 1 and −6 works.
  2. Split the middle term: 2x2 + x − 6x − 3. Group: x(2x + 1) − 3(2x + 1) = (2x + 1)(x − 3). Check by multiplying: 2x2 − 6x + x − 3 = 2x2 − 5x − 3.
  3. Set each factor to 0. 2x + 1 = 0 gives 2x = −1, so x = −1/2. x − 3 = 0 gives x = 3.
  4. Check x = −1/2: 2(1/4) − 5(−1/2) − 3 = 1/2 + 5/2 − 3 = 0. Check x = 3: 18 − 15 − 3 = 0.

Answer: Answer: x = −1/2 or x = 3

Example 5. Solve x2 + 8x + 16 = 0.

  1. The equation equals 0. Factor: product 16 and sum 8. The pair 4 and 4 works, so x2 + 8x + 16 = (x + 4)(x + 4) = (x + 4)2.
  2. Set the factor to 0: x + 4 = 0, so x = −4. The other factor is the same, so it gives the same solution.
  3. Check: (−4)2 + 8(−4) + 16 = 16 − 32 + 16 = 0.
  4. There is only one solution. It is a double root. The graph of y = x2 + 8x + 16 touches the x-axis at (−4, 0), its vertex, and stays on or above the axis.

Answer: Answer: x = −4 (a double root)

Example 6. Solve 3x2 + 21x = 0.

  1. The equation equals 0. The greatest common factor of 3x2 and 21x is 3x. Factor it out: 3x(x + 7) = 0.
  2. There are three factors: 3, x and x + 7. The number 3 is never 0, so it gives no solution.
  3. Set the others to 0: x = 0 or x + 7 = 0. So x = 0 or x = −7.
  4. Check x = 0: 3(0)2 + 21(0) = 0. Check x = −7: 3(49) + 21(−7) = 147 − 147 = 0.

Answer: Answer: x = 0 or x = −7

Example 7. A student solves x2 = 6x by dividing both sides by x and gets x = 6. What was missed? Solve it correctly.

  1. Dividing both sides by x is only allowed if x ≠ 0. The student assumed that without checking.
  2. Test x = 0 in the original equation: 02 = 6(0) is true. So x = 0 is a solution, and the student lost it.
  3. Correct method: move everything to one side, x2 − 6x = 0. Factor out the GCF x: x(x − 6) = 0.
  4. Set each factor to 0: x = 0 or x − 6 = 0. So x = 0 or x = 6.
  5. Check x = 6: 36 = 36. True.

Answer: Answer: x = 0 or x = 6. The student lost x = 0.

Example 8. Solve 4x2 − 49 = 0.

  1. The equation equals 0. Notice that 4x2 = (2x)2 and 49 = 72, so this is a difference of squares: (2x)2 − 72 = (2x − 7)(2x + 7).
  2. Set each factor to 0. 2x − 7 = 0 gives 2x = 7, so x = 7/2. 2x + 7 = 0 gives 2x = −7, so x = −7/2.
  3. Check x = 7/2: 4 · 49/4 − 49 = 49 − 49 = 0. And x = −7/2 gives the same, because (−7/2)2 = 49/4.
  4. A common slip is to answer x = ±7. That forgets the 4 in front of x2: 4(7)2 − 49 = 196 − 49 = 147, not 0.

Answer: Answer: x = 7/2 or x = −7/2

Example 9. Try to solve x2 + 2x − 4 = 0 by factoring.

  1. Look for two integers with product −4 and sum 2. The pairs with product −4 are 1 and −4, −1 and 4, and 2 and −2. Their sums are −3, 3 and 0. None equals 2.
  2. So x2 + 2x − 4 is prime. It cannot be factored using integers, and the factoring method stops here.
  3. Does the equation have a solution anyway? Look at the heights of y = x2 + 2x − 4. At x = 1, y = 1 + 2 − 4 = −1 (below the x-axis). At x = 2, y = 4 + 4 − 4 = 4 (above it). The graph is a smooth curve, so it crosses the x-axis between x = 1 and x = 2. At x = −4, y = 16 − 8 − 4 = 4, and at x = −3, y = 9 − 6 − 4 = −1, so there is another crossing between x = −4 and x = −3.
  4. So the equation has two solutions. They are not integers, so factoring with integers cannot find them. Completing the square and the quadratic formula, in later lessons, will give them exactly.

Answer: Answer: It cannot be solved by factoring with integers. It does have two solutions: one between 1 and 2, and one between −4 and −3.

Example 10. A rectangular poster is 5 inches longer than it is wide. Its area is 84 square inches. Find its width and length.

  1. Let w be the width in inches. Then the length is w + 5 inches.
  2. Area = length × width, so w(w + 5) = 84. Multiply out: w2 + 5w = 84. Subtract 84: w2 + 5w − 84 = 0.
  3. Factor: product −84 and sum 5. The pair 12 and −7 works. So (w + 12)(w − 7) = 0.
  4. Set each factor to 0: w = −12 or w = 7.
  5. Reject w = −12. A width is a length, so it cannot be negative. (It does solve the equation: (−12)(−7) = 84. It just does not describe a real poster.)
  6. So w = 7 and the length is 7 + 5 = 12. Check: 7 × 12 = 84, and 12 is 5 more than 7.

Answer: Answer: The poster is 7 inches wide and 12 inches long.

Example 11. A rectangular pool is 10 feet long and 6 feet wide. A walkway of the same width x feet runs all the way around it. The pool and walkway together cover 140 square feet. How wide is the walkway?

  1. Let x be the width of the walkway in feet. The walkway adds x on each side, so the outer rectangle is (10 + 2x) feet long and (6 + 2x) feet wide.
  2. Area of the outer rectangle: (10 + 2x)(6 + 2x) = 140. Multiply out: 60 + 20x + 12x + 4x2 = 4x2 + 32x + 60. So 4x2 + 32x + 60 = 140.
  3. Subtract 140 from both sides: 4x2 + 32x − 80 = 0. Take out the GCF 4 (or divide both sides by 4, which is not 0): x2 + 8x − 20 = 0.
  4. Factor: product −20 and sum 8. The pair 10 and −2 works. So (x + 10)(x − 2) = 0, and x = −10 or x = 2.
  5. Reject x = −10. A walkway cannot have a negative width. So x = 2.
  6. Check: the outer rectangle is 10 + 4 = 14 feet by 6 + 4 = 10 feet, and 14 × 10 = 140.

Answer: Answer: The walkway is 2 feet wide.

Example 12. Solve x2 − 4x + 3 = 0, then use the answer to describe the graph of y = x2 − 4x + 3.

  1. Factor: product 3 and sum −4. The pair −1 and −3 works. So (x − 1)(x − 3) = 0, and x = 1 or x = 3.
  2. The solutions of the equation are the x-intercepts of the parabola: the graph meets the x-axis at (1, 0) and (3, 0).
  3. The axis of symmetry is halfway between them: (1 + 3)/2 = 2. This matches the formula −b/(2a) = 4/2 = 2.
  4. The vertex is at x = 2: y = 4 − 8 + 3 = −1, so it is (2, −1). The parabola opens up (a = 1 > 0), and the vertex is below the axis, so it crosses twice. The y-intercept is c = 3, the point (0, 3).

Answer: Answer: x = 1 or x = 3. The parabola crosses the x-axis at (1, 0) and (3, 0), with vertex (2, −1).

In plain English

A quadratic equation has an x2 term in it, like x2 + 2x − 24 = 0. You are looking for the numbers that make it true.

Here is the key fact. If two numbers multiply to give 0, at least one of them is 0. So if (x + 6)(x − 4) = 0, then x + 6 = 0 or x − 4 = 0.

That only works when the product is 0. If the product is 12, or any other number, the factors could be many different things. So always get 0 on one side first.

The steps are short. Get 0 on one side. Factor completely. Set each factor to 0. Solve each one. Check your answers.

Watch for three special cases. A repeated factor gives just one solution, called a double root. A common factor of x gives the solution x = 0, so never divide both sides by x. And some quadratics do not factor with whole numbers, so you need a later method.

The solutions are also the x-intercepts of the graph. They are the places where the parabola meets the x-axis.

In a story, check each solution against the situation. A negative length makes no sense, so it gets thrown out, and you say why.

See it

Solve x² = 2x + 15 in five steps 1 Zero on one side subtract 2x and 15 from both sides x2 − 2x − 15 = 0 2 Factor completely no common factor; factor the trinomial (x + 3)(x − 5) = 0 3 Set each factor to 0 zero-product property x + 3 = 0 or x − 5 = 0 4 Solve each one two one-step equations x = −3 or x = 5 5 Check in the original use the original equation (−3)2 = 9 and 2(−3) + 15 = 9 52 = 25 and 2(5) + 15 = 25 Only a product equal to 0 splits: if A × B = 0, then A = 0 or B = 0.
The five steps for x2 = 2x + 15. Step 1 comes first because the zero-product property works only when the product equals 0.
−2−10123456 −2−1012345678 xy (1, 0)(3, 0)vertex (2, −1)(0, 3) Solve x2 − 4x + 3 = 0 Factor: (x − 1)(x − 3) = 0 Set each factor to 0 x − 1 = 0 gives x = 1 x − 3 = 0 gives x = 3 Solutions = x-intercepts The graph of y = x2 − 4x + 3 crosses the x-axis at (1, 0) and (3, 0). Halfway between 1 and 3 is x = 2, the axis of symmetry.
The solutions of x2 − 4x + 3 = 0 are the x-intercepts of y = x2 − 4x + 3: the parabola meets the x-axis at (1, 0) and (3, 0).

More ways to learn it

Fun ways and memory help

5 fun explanations (movies, sports, everyday life and more) and 6 memory tricks.

Notes and common mistakes

Key ideas, steps, mistakes to avoid and a 3-week memory plan.

Practice it your way

Flashcards

Flip, recall and space your reviews.

Cheat sheet

One page. Everything you need.

True or false

Fast yes/no questions with reasons.

Matching game

Pair them up against the clock.

Practice questions

Step-up questions with explanations.

Graphing Quadratic FunctionsFactoring TrinomialsDifference of SquaresFactoring Out the Greatest Common FactorFactoring by Grouping

How this lesson was checked

✔ AI-checked○ Teacher-checked: not yet
About standards and checking

Written by AI, then independently re-solved and checked by separate AI reviewers and automated answer checks. Written by Claude (Sonnet 5.5) from the concept template. Automated gate: schema, notation and every math answer verified by evaluated expressions. Two blind re-solves by separate AI solvers with no access to the answer key: every keyed answer matched. Two prose audits by a separate, stronger AI reviewer (Claude Opus 5.5): every finding fixed, no wrong answer keys found. Final small edits re-checked by the automated gate (no human teacher review yet). Last checked 2026-11-08.

Standards this lesson lines up with:

  • CCSS A-REI.B.4b: Solve quadratic equations by inspection (e.g., for x^2 = 49), taking square roots, completing the square, the quadratic formula and factoring, as appropriate to the initial form of the equation.
  • CCSS A-APR.B.3: Identify zeros of polynomials when suitable factorizations are available, and use the zeros to construct a rough graph of the function defined by the polynomial.
  • CCSS A-CED.A.1: Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions.