To solve a quadratic equation by factoring:
EXAMPLES:
Solve:
$x^2 = 2  x$
Solution:
Write a nice, clean list of equivalent equations:
$x^2 = 2  x$

(original equation) 
$x^2 + x  2 = 0$ 
(put in standard form: subtract $\,2\,$ from both sides; add $\,x\,$ to both sides) 
$(x+2)(x1) = 0$ 
(factor the lefthand side) 
$x+2 = 0\ \ \text{ or }\ \ x  1 = 0$ 
(use the Zero Factor Law) 
$x = 2\ \ \text{ or }\ \ x = 1$ 
(solve the simpler equations) 
Check by substituting into the original equation:
$(2)^2 \ \ \overset{\text{?}}{=}\ \ 2  (2)\,$; $\,4 = 4\,$; Check!
$(1)^2 \ \ \overset{\text{?}}{=}\ \ 2  1\,$; $\,1 = 1\,$; Check!
Solve:
$(x+3)(x2) = 0$
Solution:
Note: Don't multiply it out!
If it's already in factored form, with zero on one side,
then be happy that a lot of the work has already been done for you.
$(x+3)(x2) = 0$

(original equation) 
$x+3 = 0\ \ \text{ or }\ \ x  2 = 0$ 
(use the Zero Factor Law) 
$x = 3\ \ \text{ or }\ \ x = 2$ 
(solve the simpler equations) 
Check by substituting into the original equation:
$(3+3)(32) \ \ \overset{\text{?}}{=}\ \ 0\,$; $\,0 = 0\,$; Check!
$(2+3)(22) \ \ \overset{\text{?}}{=}\ \ 0\,$; $\,0 = 0\,$; Check!
Solve:
$(2x3)(1  3x) = 0$
Solution:
Again, don't multiply it out!
When you have a product on one side, and zero on the other side,
then you're all set to use the Zero Factor Law.
$(2x3)(1  3x) = 0$

(original equation) 
$2x3 = 0\ \ \text{ or }\ \ 1  3x = 0$ 
(use the Zero Factor Law) 
$2x = 3\ \ \text{ or }\ \ 1 = 3x$ 
(solve simpler equations) 
$\displaystyle x = \frac{3}{2}\ \ \text{ or }\ \ x = \frac{1}{3}$ 
(solve simpler equations) 
Check by substituting into the original equation:
$(2\cdot\frac323)(13\cdot\frac32)) \ \ \overset{\text{?}}{=}\ \ 0\,$; $\,0 = 0\,$; Check!
$(2\cdot\frac13+3)(13\cdot\frac13) \ \ \overset{\text{?}}{=}\ \ 0\,$; $\,0 = 0\,$; Check!
Solve:
$x^2 + 4x  5 = 0$
Solution:
Note that it's already in standard form.
$x^2 + 4x  5 = 0$

(original equation) 
$(x+5)(x1) = 0$ 
(factor the lefthand side) 
$x+5 = 0\ \ \text{ or }\ \ x  1 = 0$ 
(use the Zero Factor Law) 
$x = 5\ \ \text{ or }\ \ x = 1$ 
(solve the simpler equations) 
Check by substituting into the original equation:
$(5)^2 + 4(5)  5 \ \ \overset{\text{?}}{=}\ \ 0\,$; $25  20  5 \ \ \overset{\text{?}}{=}\ \ 0\,$; $\,0 = 0\,$; Check!
$1^2 + 4(1)  5 \ \ \overset{\text{?}}{=}\ \ 0\,$; $1 + 4  5 \ \ \overset{\text{?}}{=}\ \ 0\,$; $\,0 = 0\,$; Check!
Solve:
$14 = 5x + x^2$
Solution:
$14 = 5x + x^2$

(original equation) 
$x^2  5x  14 = 0$ 
(put in standard form: subtract $\,14\,$ from both sides; write in the conventional way) 
$(x7)(x+2) = 0$ 
(factor the lefthand side) 
$x7 = 0\ \ \text{ or }\ \ x + 2 = 0$ 
(use the Zero Factor Law) 
$x = 7\ \ \text{ or }\ \ x = 2$ 
(solve the simpler equations) 
Check by substituting into the original equation:
$14 \ \ \overset{\text{?}}{=}\ 5(7) + 7^2\,$; $14 \ \ \overset{\text{?}}{=}\ 35 + 49\,$; $\,14 = 14\,$; Check!
$14 \ \ \overset{\text{?}}{=}\ 5(2) + (2)^2\,$; $14 \ \ \overset{\text{?}}{=}\ 10 + 4\,$; $\,14 = 14\,$; Check!
Solve:
$6x = 2x^2$
Solution:
When there's no constant term, the factoring is much easier:
$6x = 2x^2$

(original equation) 
$2x^2  6x = 0$ 
(put in standard form: subtract $\,6x\,$ from both sides; write in the conventional way) 
$x^2  3x = 0$ 
(optional step: divide both sides by $\,2\,$) 
$x(x3) = 0$ 
(factor the lefthand side) 
$x = 0\ \ \text{ or }\ \ x  3 = 0$ 
(use the Zero Factor Law) 
$x = 0\ \ \text{ or }\ \ x = 3$ 
(solve the simpler equations) 
Check by substituting into the original equation:
$6\cdot 0 \ \ \overset{\text{?}}{=}\ 2\cdot 0^2\,$; $\,0 = 0\,$; Check!
$6\cdot 3 \ \ \overset{\text{?}}{=}\ 2\cdot 3^2\,$; $\,18 = 18\,$; Check!
Master the ideas from this section
by practicing the exercise at the bottom of this page.
When you're done practicing, move on to:
Factoring Trinomials
(coefficient of squared term is not $\,1\,$)
For more advanced students, a graph is displayed.
For example, the equation
$\,x^2 = 2  x\,$ is optionally accompanied
by the graph of
$\,y = x^2\,$ (the left side of the equation, dashed green)
and the graph of
$\,y=2  x\,$ (the right side of the equation, solid purple).
In this example, you are finding the values of $\,x\,$ where the green graph intersects the purple graph.
Click the “show/hide graph” button if you prefer not to see the graph.