Question 42842
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I am working on factoring trinomials. The thing is 
I am so confused with some problems. 

                   2x² + x - 6

can you help me????

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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<font color = "red">(x + 2)</font> - 3<font color = "red">(x + 2)</font>
   
We factor out the common red factor and leave 
the black factors inside parentheses:

  <font color = "red">(x + 2)</font>(2x - 3)

  (x + 2)(2x - 3)

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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<font color = "red">(2x - 5)</font> - 2<font color = "red">(2x - 5)</font>
   
We factor out the common red factor and leave 
the black factors inside parentheses:

  <font color = "red">(2x - 5)</font>(3x - 2)

  (2x - 5)(3x - 2)

Edwin</pre>