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Solving Quadratic Equations by Factoring
by Ron Kurtus (revised 18 January 2018)
One method of solving a quadratic equation is by factoring it into two linear equations and then solving each of those equations. For example, the first expression in the equation x2 + 8x + 15 = 0 can be factored into (x + 3)(x + 5), and then those two factors can then be readily solved for x.
The method requires that you first put the equation in the form of ax2 + bx + c = 0 and then try to factor ax2 + bx + c. Each factor can then be set to 0 and solved for x.
Some quadratic equations are not readily factored. In such a case, you can try to solve the equation by the completing the square method or by using the quadratic equation formula.
Questions you may have include:
- What is the general method to solve by factoring?
- What are some examples of solving by factoring?
- When is solving by factoring ineffective?
This lesson will answer those questions.
General solution
The standard form of a quadratic equation of one variable is ax2 + bx + c = 0, where:
- x is the single variable
- a, b and c are constants
- a is not equal to 0
Factoring the quadratic expression ax2 + bx + c consists of breaking the expression into two sub-expressions in the form of (dx + e)(fx + g). The quadratic equation then is:
(dx + e)(fx + g) = 0
Set each expression equal to 0 and solve them for x to get our two solutions:
dx + e = 0
dx = −e
x = −e/d
Likewise
fx + g = 0
x = −g/f
Factoring examples
Consider the quadratic equation x2 + 8x + 15 = 0.
Seeing that 3 * 5 = 15 and 3 + 5 = 8, you can factor the expression x2 + 8x + 15 into (x + 3)(x + 5). Thus, the equation is:
(x + 3)(x + 5) = 0
Since (x + 3)*0 = 0 and 0*(x + 5) = 0, you can set both expressions equal to zero and solve:
x + 3 = 0
x + 5 = 0
The solutions of the equation are:
x = −3 and x = −5
Example 2
Another example of solving by factoring is the equation:
x2 − 16 = 0
You can see that 16 = 42 and:
x2 − 42 = 0
Thus:
(x + 4)(x − 4) = 0
x + 4 = 0 and x − 4 = 0
Solutions:
x = − 4 and x = 4
or
x = ± 4
Example 3
Consider the equation:
2x2 − 3x − 14 = 0.
You can factor the expression 2x2 − 3x − 14 into (2x − 7)(x + 2).
(2x − 7)(x + 2) = 0
Thus:
2x − 7 = 0 and x + 2 = 0
The solutions are:
x = 7/2 = 3½
and
x = −2
When solving by factoring does not work
There are some quadratic equations where solving by factoring is not effective. Consider the equation
x2 − 5x + 3 = 0
You really can't factor x2 − 5x + 3 with rational numbers.
In such a case, you can try solving by the Completing the Square method or the Quadratic Formula method.
Summary
One method of solving a quadratic equation is by factoring it into two linear equations and then solving each of those equations. For example, the first expression in the equation x2 + 8x + 15 = 0 can be factored into (x + 3)(x + 5), and then those two factors can then be readily solved for x.
The method requires that you first put the equation in the form of ax2 + bx + c = 0 and then try to factor ax2 + bx + c. Each factor can then be set to 0 and solved for x.
Some quadratic equations are not readily factored. In such a case, you can try to solve the equation by the completing the square method or by using the quadratic equation formula.
Do the best you can
Resources and references
Websites
Factoring Quadratics = MathIsFun
Factoring Quadratics - PurpleMath.com
Books
Questions and comments
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Solving Quadratic Equations by Factoring