Solving Simple Quadratic Equations by Factoring

To solve a quadratic equation by factoring:

put it in standard form: $\,ax^2 + bx + c = 0\,$
factor the left-hand side
use the Zero Factor Law

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)
$(x+2)(x-1) = 0$	(factor the left-hand 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:

$(x+3)(x-2) = 0$

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.

$(x+3)(x-2) = 0$	(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:

$(-3+3)(-3-2) \ \ \overset{\text{?}}{=}\ \ 0\,$; $\,0 = 0\,$; Check!

$(2+3)(2-2) \ \ \overset{\text{?}}{=}\ \ 0\,$; $\,0 = 0\,$; Check!

Solve:

$(2x-3)(1 - 3x) = 0$

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.

$(2x-3)(1 - 3x) = 0$	(original equation)
$2x-3 = 0\ \ \text{ or }\ \ 1 - 3x = 0$	(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:

$(2\cdot\frac32-3)(1-3\cdot\frac32)) \ \ \overset{\text{?}}{=}\ \ 0\,$; $\,0 = 0\,$; Check!

$(2\cdot\frac13+3)(1-3\cdot\frac13) \ \ \overset{\text{?}}{=}\ \ 0\,$; $\,0 = 0\,$; Check!

Solve:

$x^2 + 4x - 5 = 0$

Solution:
Note that it's already in standard form.

$x^2 + 4x - 5 = 0$	(original equation)
$(x+5)(x-1) = 0$	(factor the left-hand 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)
$(x-7)(x+2) = 0$	(factor the left-hand side)
$x-7 = 0\ \ \text{ or }\ \ x + 2 = 0$	(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 $\,2\,$)
$x(x-3) = 0$	(factor the left-hand 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.