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Math
A |
Factoring
Trinomials
(a 
1) |
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for this lesson
a will
NOT be 1.
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In
the previous lesson on factoring trinomials where a = 1, we learned that
factoring requires that we put our investigative skills to work.
Those skills will really be put to the test when our trinomial starts
with a number other than one.
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When
the leading coefficient is a number other than one, the
number of possible answers increases ..... making our
investigative efforts harder. |
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Let's see what
is involved with factoring 
| 1. |
First, check to see if all
of the terms share a common
factor which may be removed. If each term can be
factored before you begin, your work will be easier. The
terms in this problem do not have a common factor.
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| 2. |
Consider all
of the possible factors of the leading coefficient, 2x². In
this problem we only have one choice, 2x and x. So we can
start with:
(2x
) (x )
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| 3. |
Consider all
of the possible factors of the last term, -6. The possible
answers are:
+6 and -1
-6 and +1
+3 and -2
-3 and +2 |
you
need to consider
all of the possible
ways of obtaining
the number -6 |
This tells us that our possible answers to
this problem will be:
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(2x + 6)(x - 1)
(2x - 6)(x + 1)
(2x +
3)(x - 2)
(2x - 3)(x + 2) |
(x + 6)(2x - 1)
(x - 6)(2x + 1)
(x + 3)(2x - 2)
(x - 3)(2x + 2) |
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| 4.
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You need to test each of these pairs to see which will yield
the correct middle term.
(2x + 6)(x - 1) gives
middle term 4x.
(2x - 6)(x + 1) gives
middle term -4x.
(2x + 3)(x - 2) gives
middle term -x. YEA!!!!!
(2x - 3)(x + 2) gives
middle term +x.
(x + 6)(2x - 1) gives
middle term 11x.
(x - 6)(2x + 1) gives
middle term -11x.
(x + 3)(2x - 2) gives
middle term 4x.
(x - 3)(2x + 2) gives
middle term -4x.
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| 5. |
Answer:
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Unfortunately,
there are often many possible answers that need to be
considered.
But don't let it drive you crazy ... just slowly and
systematically examine your possible answers until you find
the one that yields the correct middle term.
You can do it!!! |
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