Solving Simple Quadratic Equations by Factoring

Need some basic practice with quadratic equations first?

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!

Concept Practice

For more advanced students, a graph is available. 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 to toggle the graph.