Factor the following polynomial completely showing all steps:

Question

6x^2-23x-55

Answer 1

For any ax² + bx + c  there is a process that always works.

 

First, find ac.   For this problem a=6, b = -23,  c = -55.  So ac = (6)(-55) = -330

second, find the factors of ac   -330 = (-3)(110)  do these add to b (-23)?  NO, they add to 107 which is not b, so not correct.

                                             -330 = (-10)(33)  do these add to b (-23)?  NO, they add to +23, so if we changed the signs around, we would have found the two numbers we want.

 

                                             -330 = (10)(-33)  These add to -23, so we move onto Step 3.

 

Step 3:  Replace "b" with the factors, meaning,  we will replace -23x with  (10x) + (-33x)  and keep everything else the same.

 

6x² + 10x + (-33x) - 55   Notice this is equivalent to the original problem if you combine the like terms 10x and -33x.

 

Now group the first 2 terms together, and group the last 2 terms together.

 

(6x² +10x) + (-33x - 55)   then factor out of each group whatever will factor out

 

2x(3x + 5) + (-11)(3x + 5)      

NOTICE  both groups had a (3x+5) factor.  This is the indicator that you are doing it right!

 

Factor the (3x+5) out of each group:   (3x+5)(2x - 11)  <-- it is now factored completely!

 

To check, re "FOIL" it:   (3x)(2x) + (3x)(-11) + (5)(2x) + (5)(-11)

 

  = 6x² -33x +10x - 55    Combine like terms and get  6x² -23x -5   this is the original problem, so we factored it correctly.

 

The final answer then is   (3x + 5)(2x - 11)

 

Answer 2

Fortunately for us, 55 has only two factors. So we have

(  5)(   11)

with pluses or minuses in front of the 5 and 11, and 6x² factored as either 6x•x or 3x•2x.

Trying 3x and 2x first gives (3x+5)(2x-11)

 

Wow, lucky guess! So it is with factoring like this.