# SOLUTION: 6x^2+18x-600 Factor, if possible. Be sure to factor completely. Always factor out the greatest commone factor first, if one exist. thanks.

Algebra ->  -> SOLUTION: 6x^2+18x-600 Factor, if possible. Be sure to factor completely. Always factor out the greatest commone factor first, if one exist. thanks.       Log On

 Ad: Mathway solves algebra homework problems with step-by-step help! Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!

Question 725322: 6x^2+18x-600
Factor, if possible. Be sure to factor completely. Always factor out the greatest commone factor first, if one exist.
thanks.

You can put this solution on YOUR website!

Factor out the GCF .

Now let's try to factor the inner expression

---------------------------------------------------------------

Looking at the expression , we can see that the first coefficient is , the second coefficient is , and the last term is .

Now multiply the first coefficient by the last term 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 :
1,2,4,5,10,20,25,50,100
-1,-2,-4,-5,-10,-20,-25,-50,-100

Note: list the negative of each factor. This will allow us to find all possible combinations.

These factors pair up and multiply to .
1*(-100) = -100
2*(-50) = -100
4*(-25) = -100
5*(-20) = -100
10*(-10) = -100
(-1)*(100) = -100
(-2)*(50) = -100
(-4)*(25) = -100
(-5)*(20) = -100
(-10)*(10) = -100

Now let's add up each pair of factors to see if one pair adds to the middle coefficient :

First NumberSecond NumberSum
1-1001+(-100)=-99
2-502+(-50)=-48
4-254+(-25)=-21
5-205+(-20)=-15
10-1010+(-10)=0
-1100-1+100=99
-250-2+50=48
-425-4+25=21
-520-5+20=15
-1010-10+10=0

From the table, we can see that there are no pairs of numbers which add to . So cannot be factored.

===============================================================