Factoring a Quadratic Trinomial by Grouping
Another method for factoring
these kinds of quadratic trinomials
is called factoring by grouping.
Factoring by grouping can
be a bit more tedious, and
is often not worth the trouble
if you can find the correct
factors by some quick trial
and error. However, it works
quite well when the factors
are not immediately obvious,
such as when you have a very
large number of candidate
factors. When this happens,
the trial and error method
becomes very tedious
Factoring by grouping is
best demonstrated with a few
examples.
Example:
|
Given: |
5x2 + 11x + 2 |
|
Find
the product ac: |
(5)(2) = 10 |
|
Think
of two factors of 10
that add up to 11: |
1 and 10 |
|
Write
the 11x as the
sum of 1x and
10x: |
5x2 + 1x + 10x + 2 |
|
Group
the two pairs of terms: |
(5x2
+ 1x) + (10x + 2) |
|
Remove
common factors from
each group: |
x(5x + 1) + 2(5x + 1) |
|
Notice
that the two quantities
in parentheses are now
identical. That means
we can factor out a
common factor of (5x + 1): |
(5x + 1)(x + 2) |
Example:
|
Given: |
4x2 + 7x 15 |
|
Find
the product ac: |
(4)(-15) = -60 |
|
Think
of two factors of -60 that add up to 7: |
-5
and 12 |
|
Write
the 7x as the
sum of -5x and 12x: |
4x2 5x + 12x 15 |
|
Group
the two pairs of terms: |
(4x2 5x) + (12x 15) |
|
Remove
common factors from
each group: |
x(4x 5) + 3(4x 5) |
|
Notice
that the two quantities
in parentheses are now
identical. That means
we can factor out a
common factor of (4x - 5): |
(4x 5)(x + 3) |
Given a general quadratic
trinomial
ax2 + bx + c
1. Find the product ac.
2. Find two numbers h
and k such that
hk = ac
(h
and k are factors of
the product of the coefficient
of x2 and
the constant term)
AND
h + k = b
(h
and k add to give the
coefficient of x)
3. Rewrite the quadratic
as
ax2 + hx + kx + c
4. Group the two pairs of
terms that have common factors
and simplify.
(ax2 + hx) + (kx + c)
x(ax + h) + (kx + c)
(note:
because of the way you chose
h and k, you
will be able to factor a constant
out of the second parentheses,
leaving you with two identical
expressions in parentheses
as in the examples).
·
Remember that
this wont work for all quadratic
trinomials, because not all
quadratic trinomials can be
factored into products of
binomials with integer coefficients.
If you have a non-factorable
trinomial, you will not be
able to do step 2 above.
|
Suppose the quadratic
trinomial in question
came from multiplying
two arbitrary binomials:
(px + n)(qx + m)
If we multiply this
out we will get
pqx2 + pmx + qnx + nm
or
pqx2 + (pm + qn)x + nm
Notice that the coefficient
of x consists
of a sum of two terms,
pm and qn.
These are the two numbers
we called h and
k above.
pm = h
qn = k
Now we see that the
two numbers h
and k add up
to the coefficient of
x, which we called
b:
h + k = b
Obviously they are
factors of their own
product pmqn,
but we notice that pq = a,
and mn = c,
so
(pm)(qn) = (pq)(nm)
which is equivalent
to
hk = ac |
|
