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Algbrea 2 Math

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2 Answers

David, thanks for your question. The two trinomials you used as examples are very similar : ax2+bx+c is the general form and x2+bx+c is the specific case where a = 1.

when c=0: ax2+bx+c --> ax2+bx
There is a common factor of x in both terms so x(ax+b) = ax2+bx

Example: 5x2-4x = x(5x-4) (a=5, b=-4, c=0)

when b=0 and c<0: ax2+bx+c --> ax2+c
let d = -c. d>0 because c<0 and ax2+c = ax2-d
The difference of two squares --> (x√a + √d)(x√a - √d) = ax2-d
d = -c, so (x√a + √(-c))(x√a - √(-c)) = ax2+c
Remember that this only works when b=0 and c<0 and can only be factored if a is a
perfect square and if -c is a perfect square.

Example: 9x2-25 = (x√9 + √(25))(x√9 - √(25)) = (3x + 5)(3x - 5)
or (-3x + 5)(-3x - 5)
(a=9, b=0, c=-25)

when a=1: ax2+bx+c --> x2+bx+c
There are a number of ways to approach factoring a trinomial in this form, but
no matter the technique you need a result which looks like (x + d)(x + e)

You need to find a d and e such that (d)(e) = c AND dx + ex = bx, or d + e = b
Not every trinomial can be factored!

Example: x2-3x-28 (a=1, b=-3, c=-28)
need to find d and e such that (d)(e) = -28 and d + e = -3
Since c is negative, then d and e will be a pair with one positive and
one negative.
The factors of 28 are 1&28, 2&14, 4&7. For -28 one of the pair is negative.
We need to use one of these pairs as d and e with one of them negative.
Using d=4 and e=-7, (4)(-7) = -28 and 4+(-7)=-3
(x + d)(x + e) = (x + 4)(x - 7) = x2-3x-28

When a=-1: ax2+bx+c --> -x2+bx+c
Same idea, but your result will look like (x + d)(-x + e)
You need to find a d and e such that (d)(e) = c AND ex - dx = bx, or e-d = b

Example: -x2+2x+15 (a=-1, b=2, c=15)
The factors of 15 are 1&15, 3&5, -1&-15, -3&-5.
We need to use one of these pairs as d and e.
Use d=3 and e=5, (3)(5)=15, 5-3 = 2 --> (x+3)(-x+5)
OR use d=-5 and e=-3, (-5)(-3)=15, -3 -(-5) = 2 --> (x-5)(-x-3)

The general case ax2+bx+c: This is more complex than x2+bx+c because you must now find two more
constants, f and g so that your result will be (fx + d)(gx +e)
If you expand this form you get (fg)x2 + (dg +ef)x +de
This requires that (d)(e) = c AND (f)(g) = a, AND dg +ef = b

Example: 8x2+2x -21 (a=8, b=2, c=-21)
Factors of 21 are 1&21, 3&7. -21 means one + one -
Factors of 8 are 1&8, 2&4. Need both + or both -
Since b is small, experience tells me to try combinations of 3&7, and 2&4
Try d=3, e=-7, f=2, g=4.
3(-7)=-21, 2(4) = 8, (3)(4)+(-7)(2)=-2 (needs to =2) DOESN'T WORK
Try swapping signs for d and e, d=-3, e=7, f=2, g=4
-3(7)=-21, 2(4)=8, -3(4)+7(2)=2 WORKS!
So, 8x2+2x -21 = (fx+d)(gx+e) = (2x-3)(4x+7)
 Idea behind it is, a quadratic   aX+ bx + c is factorable  to ( dx + m ) ( x + n) , if and only if there exist 3
   integer d, m , n such that:
 
     aX2   + bx + c = ( dX +m) ( x + n)
 
 
     Give an example to clear:
 
      2 X + 23 X + 38    , we want to see factor ability of this quadratic.
   
          Have to look at, 2 numbers whose product is 2*38 = 76, and their sum is 23
 
         Like the system of equation  :
 
         m+ n = 23
         mn = 76
        has integer solution, so solve and find out.
 
       An easy short cut will be:
        Write down  76 as product of its factors, same way as done in arithmetic using division tree.
 
         76 = 2 * 38 = 4 * 19 
 
           Now we see that  4 + 19 = 23 , and that is what we were looking for and the quadratic is factorable.
 
      Next step is to break down 23 X  to   23X = 19X + 4X
 
 
      2X2 + 4X + 19X + 38 
 
       Now we follow by factoring by group.
 
     2X ( X+ 2) 19( X +2)= ( X+2) ( 2X +1 9)      is the final answer.
      
     There are multiple of ways to find, like trial and error of trying to multiply 2 factors and see when becomes equal to Sum.
   But all those methods stem from the same principle. Our focus should be on the principle where the methods stem from, rather than getting the answer.