6.6: Solving Quadratic Equations by Factoring
- Page ID
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| Title | Level of Approach | Type | Length |
|---|---|---|---|
| Solving Quadratic Equations Using the Zero Factor Principle | Elementary Algebra | Review | 13:53 |
| An Introduction to Quadratic Equations and Extraction of Roots | Elementary Algebra | Lecture | 52:24 |
| Solving Quadratic Equations by Completing the Square | Elementary Algebra | Lecture | 37:28 |
| Solving Quadratic Equations by Factoring | Elementary Algebra | Lecture | 22:21 |
| Proof of the Quadratic Formula | Intermediate Algebra | Proof | 11:41 |
| Solving Quadratic Equations Using the Quadratic Formula | Intermediate Algebra | Lecture | 50:34 |
Definitions and Theorems
You have already learned to factor quadratic expressions. Here you pair that skill with a single property of the real numbers to solve quadratic equations.
A quadratic equation is an equation that can be written in the standard form\[ax^2 + bx + c = 0,\nonumber\]where \(a\), \(b\), and \(c\) are real numbers and \(a \neq 0\).
For real numbers \(a\) and \(b\), if \(ab = 0\), then \(a = 0\) or \(b = 0\) (or both).
To solve a quadratic equation by factoring, work in three moves: write the equation in standard form so that one side is \(0\); factor the nonzero side completely; then apply the Zero Product Property, setting each factor equal to \(0\) and solving the resulting linear equations.
The Zero Product Property applies only when the product equals \(0\). An equation such as \((x-2)(x+3) = 5\) cannot be solved by setting each factor equal to \(5\); you must first rewrite the equation so that one side is \(0\).
Examples
Solve \((2x-1)(x+4) = 0\).
- Answer
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The left side is already a product of two factors equal to \(0\), so apply the Zero Product Property and set each factor equal to \(0\):\[2x - 1 = 0 \quad \text{or} \quad x + 4 = 0.\nonumber\]Solving each linear equation gives\[x = \dfrac{1}{2} \quad \text{or} \quad x = -4.\nonumber\]
Solve \(x^2 + 3x - 10 = 0\).
- Answer
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The equation is already in standard form. Factor the trinomial by finding two numbers whose product is \(-10\) and whose sum is \(3\); these are \(5\) and \(-2\):\[(x+5)(x-2) = 0.\nonumber\]By the Zero Product Property,\[x + 5 = 0 \quad \text{or} \quad x - 2 = 0,\nonumber\]so \(x = -5\) or \(x = 2\).
Solve \(x^2 = 7x - 12\).
- Answer
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First write the equation in standard form by moving every term to one side:\[\begin{aligned} x^2 &= 7x - 12 \\ x^2 - 7x + 12 &= 0. \end{aligned}\nonumber\]Factor by finding two numbers whose product is \(12\) and whose sum is \(-7\); these are \(-3\) and \(-4\):\[(x-3)(x-4) = 0.\nonumber\]By the Zero Product Property, \(x - 3 = 0\) or \(x - 4 = 0\), so \(x = 3\) or \(x = 4\).
Solve \(6x^2 + x - 2 = 0\).
- Answer
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The equation is in standard form with a leading coefficient other than \(1\). Multiply the leading coefficient and the constant term: \((6)(-2) = -12\). Find two numbers whose product is \(-12\) and whose sum is the middle coefficient \(1\); these are \(4\) and \(-3\). Rewrite the middle term and factor by grouping:\[\begin{aligned} 6x^2 + x - 2 &= 0 \\ 6x^2 + 4x - 3x - 2 &= 0 \\ 2x(3x+2) - (3x+2) &= 0 \\ (3x+2)(2x-1) &= 0. \end{aligned}\nonumber\]By the Zero Product Property, \(3x + 2 = 0\) or \(2x - 1 = 0\), so\[x = -\dfrac{2}{3} \quad \text{or} \quad x = \dfrac{1}{2}.\nonumber\]
Solve \(x^2 - 8x + 16 = 0\).
- Answer
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Factor the trinomial; two numbers whose product is \(16\) and whose sum is \(-8\) are \(-4\) and \(-4\), so the trinomial is a perfect square:\[(x-4)^2 = 0.\nonumber\]The single repeated factor gives \(x - 4 = 0\), so \(x = 4\). Because the factor \(x - 4\) occurs twice, \(x = 4\) is called a double root.
Solve \(5x^2 = 15x\).
- Answer
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Move every term to one side; do not divide both sides by \(x\) (see the Caution below):\[\begin{aligned} 5x^2 &= 15x \\ 5x^2 - 15x &= 0. \end{aligned}\nonumber\]Factor out the common factor \(5x\):\[5x(x-3) = 0.\nonumber\]By the Zero Product Property, \(5x = 0\) or \(x - 3 = 0\), so \(x = 0\) or \(x = 3\).
When both sides of an equation share a factor of the variable, resist the urge to divide it out. Dividing \(5x^2 = 15x\) by \(x\) yields \(5x = 15\), giving only \(x = 3\) and silently discarding the solution \(x = 0\). Always move every term to one side and factor instead.
Sources
Several parts of this text use modifications from the following sources:
- Wikipedia article: "Zero-product property"
- Wikipedia article: "Quadratic equation"
All of these sources are released under the Creative Commons Attribution-Share-Alike License 4.0.