Question 422701: This is on factoring completely. Most of them I have been able to do, but this one I am stuck on. I can't find a GCF for all the terms. According to the back of the book it is factorable. How would I completely factor this:
a^2-8ab-33b^2
Thanks
Found 2 solutions by rfer, jim_thompson5910:
Answer by rfer(16322)   (Show Source):
Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website! Looking at the expression  , we can see that the first coefficient is  , the second coefficient is  , and the last coefficient is  .
Now multiply the first coefficient by the last coefficient to get  .
Now the question is: what two whole numbers multiply to (the previous product) and add to the second coefficient ?
To find these two numbers, we need to list all of the factors of (the previous product).
Factors of :
1,3,11,33
-1,-3,-11,-33
Note: list the negative of each factor. This will allow us to find all possible combinations.
These factors pair up and multiply to .
1*(-33) = -33
3*(-11) = -33
(-1)*(33) = -33
(-3)*(11) = -33
Now let's add up each pair of factors to see if one pair adds to the middle coefficient :
| First Number | Second Number | Sum | | 1 | -33 | 1+(-33)=-32 | | 3 | -11 | 3+(-11)=-8 | | -1 | 33 | -1+33=32 | | -3 | 11 | -3+11=8 |
From the table, we can see that the two numbers  and  add to (the middle coefficient).
So the two numbers and both multiply to and add to
Now replace the middle term  with  . Remember, and add to . So this shows us that  .
 Replace the second term with .
 Group the terms into two pairs.
 Factor out the GCF  from the first group.
 Factor out the GCF  from the second group.
 Factor out  from the entire expression.
So completely factors to
If you need more help, email me at jim_thompson5910@hotmail.com
Also, please consider visiting my website: http://www.freewebs.com/jimthompson5910/home.html and making a donation. Thank you
Jim
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