Question 42842: I am working on factoring trinomials. The thing is I am so confused with some problems.
2x² + x - 6
can you help me????
Found 3 solutions by josmiceli, fractalier, AnlytcPhil:
Answer by josmiceli(19441)   (Show Source):
Answer by fractalier(6550)  (Show Source):
You can put this solution on YOUR website! Okay from
2x^2 + x - 6
what we do is to set up our binomial's parentheses first
( )( )
then we look at the second sign, the minus in front of the 6...
since it's minus, we know our binomial factors have different signs...
( + )( - )
then we say "what are the factors of 12 (the 2 times the 6) that have a difference of 1 (the coefficient of the middle term)?"
The factors are 4 and 3...we then try to place the factors of 2x^2 and the factors of -6 in such a way that when we FOIL it out we get the +1x middle term...we get
(x + 2)(2x - 3)
Answer by AnlytcPhil(1806)  (Show Source):
You can put this solution on YOUR website!
I am working on factoring trinomials. The thing is
I am so confused with some problems.
2x² + x - 6
can you help me????
-------------------------------------------------
To factor Ax² ± Bx ± C when there is no common factor
of |A|, |B|, and |C| and A is positive.
1. Multiply |A| by |C|, getting AC
2. If the last sign is +, think of two positive integers
which have product AC and which have SUM |B|
If the last sign is -, think of two positive integers
which have product AC and which have DIFFERENCE |B|
3. Rewrite the middle term ±Bx using those two integers
found in the preceding step, attaching appropriate
signs.
4. Factor by grouping.
In your problem, 2x² + 1x - 6, I placed the 1 coefficient
beside the middle term for emphasis.
1. Multiply |A| by |C|, getting AC
Multiply |2| by |-6| or 2×6 getting 12
2. The last sign is -, so think of two positive integers
which have product AC and which have DIFFERENCE |B|
So we think of two positive integers which have product
12 and difference of 1. These are 4 and 3, because
4×3=12 and 4-3 = 1
3. Rewrite the middle term ±Bx using those two integers
found in the preceding step, attaching appropriate \
signs.
We rewrite +1x as +4x - 3x. So now we have
2x² + 4x - 3x - 6
4. Factor by grouping.
2x² + 4x - 3x - 6
Factor 2x out of the first two terms
2x(x + 2) - 3x - 6
Factor -3 out of the last two terms. Notice
that you factor out a negative when the next
to the last term is preceded by a minus sign.
Also when factoring out a nagative, the sign
of the last term changes:
2x(x + 2) - 3(x + 2)
Notice the common factor (x + 2) which I will
color red for emphasis:
2x(x + 2) - 3(x + 2)
We factor out the common red factor and leave
the black factors inside parentheses:
(x + 2)(2x - 3)
(x + 2)(2x - 3)
--------------------------
Here is another example:
Factor 6x² - 19x + 10.
1. Multiply |A| by |C|, getting AC
Multiply |6| by |10| or 6×10 getting 60
2. The last sign is +, so think of two positive integers
which have product AC and which have SUM |B|
So we think of two positive integers which have product
60 and sum of 19. These are 15 and 4, because
15×4=60 and 15+4 = 19
3. Rewrite the middle term ±Bx using those two integers
found in the preceding step, attaching appropriate signs.
We rewrite -19x as -15x - 4x. So now we have
6x² - 15x - 4x + 10
4. Factor by grouping.
6x² - 15x - 4x + 10
Factor 3x out of the first two terms
3x(2x - 5) - 4x + 10
Factor -2 out of the last two terms. Notice
that you factor out a negative when the next
to the last term is preceded by a minus sign.
Also when factoring out a nagative, the sign
of the last term changes:
3x(2x - 5) - 2(2x - 5)
Notice the common factor (2x - 5) which I will
color red for emphasis:
3x(2x - 5) - 2(2x - 5)
We factor out the common red factor and leave
the black factors inside parentheses:
(2x - 5)(3x - 2)
(2x - 5)(3x - 2)
Edwin
|
|
|
