Unit 6 · Lesson 6 of 6
Solving equations by factoring
You'll be able to: solve an equation like \(x^2 + 2x = 15\) by getting 0 on one side, factoring, and setting each factor equal to 0.
Step 1 of 5Warm-up
Two quick questions from earlier lessons. You will use both skills today.
Factor completely: \(3x^2 - 6x - 24\).
GCF first: \(3(x^2 - 2x - 8)\).
Inside: two numbers that multiply to \(-8\) and add to \(-2\) are \(-4\) and \(2\).
Answer: \(3(x - 4)(x + 2)\). Stopping at \(3(x^2 - 2x - 8)\) is not complete.
Solve \(5x - 3 = 2(x + 6)\).
Distribute: \(5x - 3 = 2x + 12\).
Subtract \(2x\) and add 3: \(3x = 15\). Divide by 3: \(x = 5\).
Step 2 of 5The idea
If two numbers multiply to 0, at least one of them must be 0. There is no other way to get 0 from multiplying. This is the zero product property.
So if \((x - 3)(x + 5) = 0\), then \(x - 3 = 0\) or \(x + 5 = 0\). That gives two solutions: \(x = 3\) or \(x = -5\).
Solving by factoring
1. Move everything to one side so the other side is 0. For a quadratic this is standard form, \(ax^2 + bx + c = 0\).
2. Factor completely.
3. Set each factor with an \(x\) equal to 0 and solve each small equation.
4. Check each solution in the original equation.
Never divide both sides by \(x\). If \(x = 0\) is a solution, dividing by \(x\) throws it away. Factor out the \(x\) instead.
Why this shows up on the placement test: solving quadratic equations is one of the core skills the test checks, and factoring is the fastest method when it works.
Step 3 of 5Worked examples
Example 1: Solve \(x^2 + 2x = 15\)
- Get 0 on one side: subtract 15 from both sides. \(x^2 + 2x - 15 = 0\).
- Factor: two numbers that multiply to \(-15\) and add to 2 are \(5\) and \(-3\). So \((x + 5)(x - 3) = 0\).
- Set each factor to 0: \(x + 5 = 0\) gives \(x = -5\); \(x - 3 = 0\) gives \(x = 3\).
- Check \(x = 3\): \(9 + 6 = 15\). Check \(x = -5\): \(25 - 10 = 15\). Both work. Solutions: \(x = -5\) or \(x = 3\).
Example 2: Solve \(4x^2 = 12x\)
- Get 0 on one side: \(4x^2 - 12x = 0\). Don't divide by \(x\). That would lose a solution.
- Factor out the GCF: \(4x(x - 3) = 0\).
- Set each factor with an \(x\) to 0: \(4x = 0\) gives \(x = 0\); \(x - 3 = 0\) gives \(x = 3\).
- Check \(x = 0\): \(0 = 0\). Check \(x = 3\): \(4 \cdot 9 = 36\) and \(12 \cdot 3 = 36\). Solutions: \(x = 0\) or \(x = 3\).
Step 4 of 5Practice
Type all the solutions, separated by commas, and press Check. If you miss twice, open the worked solution. Then mark how it went.
Solve \((x - 4)(x + 7) = 0\).
It is already factored with 0 on one side.
\(x - 4 = 0\) gives \(x = 4\); \(x + 7 = 0\) gives \(x = -7\).
Answer: \(x = 4\) or \(x = -7\).
Solve \(x^2 - 5x - 14 = 0\).
Two numbers that multiply to \(-14\) and add to \(-5\): \(-7\) and \(2\). So \((x - 7)(x + 2) = 0\).
Answer: \(x = 7\) or \(x = -2\).
Solve \(2x^2 + 10x = 0\).
Factor out the GCF \(2x\): \(2x(x + 5) = 0\).
\(2x = 0\) gives \(x = 0\); \(x + 5 = 0\) gives \(x = -5\).
Answer: \(x = 0\) or \(x = -5\). Dividing by \(x\) at the start would have lost \(x = 0\).
Solve \(x^2 = 8x - 16\).
Get 0 on one side: \(x^2 - 8x + 16 = 0\).
Perfect square trinomial: \((x - 4)^2 = 0\), which is \((x - 4)(x - 4) = 0\).
Both factors give the same value. Answer: \(x = 4\) (only one solution).
Solve \(2x^2 + x = 6\).
Get 0 on one side: \(2x^2 + x - 6 = 0\).
ac method: \(ac = -12\); multiply to \(-12\), add to 1: \(4\) and \(-3\). Split: \(2x^2 + 4x - 3x - 6 = 2x(x + 2) - 3(x + 2) = (x + 2)(2x - 3)\).
\(x + 2 = 0\) gives \(x = -2\); \(2x - 3 = 0\) gives \(2x = 3\), so \(x = \dfrac{3}{2}\).
Answer: \(x = -2\) or \(x = \dfrac{3}{2}\).
Solve \(9x^2 - 4 = 0\).
Difference of squares: \((3x + 2)(3x - 2) = 0\).
\(3x + 2 = 0\) gives \(x = -\dfrac{2}{3}\); \(3x - 2 = 0\) gives \(x = \dfrac{2}{3}\).
Answer: \(x = \dfrac{2}{3}\) or \(x = -\dfrac{2}{3}\).
Solve \(x(x + 3) = 10\).
One side is 10, not 0, so multiply out and subtract 10: \(x^2 + 3x - 10 = 0\).
Two numbers that multiply to \(-10\) and add to 3: \(5\) and \(-2\). So \((x + 5)(x - 2) = 0\).
Answer: \(x = -5\) or \(x = 2\). Check \(x = 2\): \(2 \cdot 5 = 10\).
A rectangle's length is 3 meters more than its width. Its area is 40 square meters. What is the width, in meters?
Let the width be \(w\). Then the length is \(w + 3\), and area = length × width: \(w(w + 3) = 40\).
Get 0 on one side: \(w^2 + 3w - 40 = 0\). Multiply to \(-40\), add to 3: \(8\) and \(-5\). So \((w + 8)(w - 5) = 0\).
\(w = -8\) or \(w = 5\). A width can't be negative, so \(w = 5\).
Answer: 5 meters. Check: the length is 8, and \(5 \cdot 8 = 40\).
Step 5 of 5Wrap-up
You can now solve an equation by factoring: get 0 on one side, factor completely, and set each factor equal to 0. You also know not to divide by \(x\).
Read or watch next: OpenStax Intermediate Algebra 2e, 6.5 Polynomial Equations (free textbook, with word problems and more practice).
Also good: Khan Academy: Solving quadratics by factoring (short videos). Quick rules: Factoring cheat sheet.
Something unclear? Ask your teacher bot a follow-up question, for example: "Why can't I divide both sides of 4x² = 12x by x?" It knows this course and can explain things another way.