SOLVING SIMPLE QUADRATIC EQUATIONS BY FACTORING

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)
$(x+2)(x-1) = 0$ (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.

CONCEPT QUESTIONS EXERCISE:
On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
PROBLEM TYPES:
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AVAILABLE MASTERED IN PROGRESS

Solve: