SOLUTION: 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 fact

Algebra.Com
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): You can put this solution on YOUR website!
(a+3b)(a-11b)
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 NumberSecond NumberSum
1-331+(-33)=-32
3-113+(-11)=-8
-133-1+33=32
-311-3+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

RELATED QUESTIONS

I have been reviewing factoring trinomials. I am kind of stuck on this problem. I got an... (answered by MathLover1)
Hi I am stuck on this one which has me completely bamboozled. Your assistance would be... (answered by solver91311,stanbon)
Factor completely: x^10y^3 – 4x^9y^2 – 21x^8y If you believe that a polynomial is not (answered by stanbon)
I am working on a problem that I am having difficulty answering. There is not an answer... (answered by psbhowmick,richwmiller)
I am having a difficult time with this question, its been 10 years since needing to do... (answered by solver91311)
Hi Again, Thanks for everyone helping me! I am stuck on this one! Here is the... (answered by stanbon)
I am in college Prealgebra and Introductory Algebra. This Greek to me and I seriously... (answered by irla,ewatrrr,lenny460)
I am just beginning sixth grade math team, and I am stuck on this problem in summer... (answered by Alan3354)
Figuring out graphing of this polynomial: f(x) = 12x^3 - 12x^2 - 24x what I have... (answered by josgarithmetic)