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Solving a
quadratic (or any kind of
equation) by factoring it
makes use of a principle known
as the zero-product rule.
Zero Product Rule
If
ab = 0
then either a = 0
or b = 0
(or both).
In
other words, if the
product of two things
is zero then one of
those two things must
be zero, because the
only way to multiply
something and get zero
is to multiply it by
zero.
Thus,
if you can factor an
expression that is
equal to zero, then
you can set each factor
equal to zero and solve
it for the unknown.
·
The
expression must
be set equal to zero
to use this principle
·
You
can always make any
equation equal to zero
by moving all the terms
to one side.
Example:
|
Given: |
x2 – x = 6 |
|
Move all terms to one side: |
x2 – x – 6 = 0 |
|
Factor: |
(x – 3)(x + 2) = 0 |
|
Set each factor equal to zero and solve: |
(x – 3) = 0 OR (x + 2) = 0 |
|
Solutions: |
x
= 3 OR
x = -2
|
|
If
a quadratic equation has no
constant term (i.e. c = 0)
then it can easily be solved
by factoring out the common
x from the remaining
two terms:
Then,
using the zero-product rule,
you set each factor equal
to zero and solve to get the
two solutions:
x = 0
or ax + b = 0
WARNING:
Do not divide out the common
factor of x or you
will lose the x = 0
solution. Keep all the factors
and use the zero-product rule
to get the solutions.
When a quadratic has all
three terms, you can still
solve it with the zero-product
rule if you are able to factor
the trinomial.
·
Remember,
not all trinomial quadratics
can be factored with
integer constants
If it can be factored, then
it can be written as a product
of two binomials. The zero-product
rule can then be used to set
each of these factors equal
to zero, resulting in two
equations that are both simple
linear equations that can
be solved for x. See
the above example for the
zero-product rule to see how
this works.
A more thorough discussion
of factoring trinomials may
be found in the chapter on
polynomials, but here is a
quick review:
1.
Clear fractions (by
multiplying through by the
common denominator)
2.
Remove common factors
if possible
3.
If the coefficient
of the x2
term is 1, then
x2 + bx + c = (x + n)(x + m),
where n and m
i.
Multiply to give c
ii.
Add to give b
4.
If the coefficient
of the x2
term is not 1, then use either
a. Guess-and
Check
i.
List the factors of
the coefficient of the x2
term
ii.
List the factors of
the constant term
iii.
Test all the possible
binomials you can make from
these factors
b. Factoring
by Grouping
i.
Find the product ac
ii.
Find two factors
of ac that add to give
b
iii.
Split the middle
term into the sum of two terms,
using these two factors
iv.
Group the terms into
pairs
v.
Factor out the common
binomial |
