Question 147455This question is from textbook algebra
: This is factoring pattern for ax^2+bx+c. The problem is 3x^2+11x+10, the answer I got was (3x+2)(x-5), Thanks for the help=)
This question is from textbook algebra
Found 2 solutions by jim_thompson5910, vleith:
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
Looking at  we can see that the first term is  and the last term is  where the coefficients are 3 and 10 respectively.
Now multiply the first coefficient 3 and the last coefficient 10 to get 30. Now what two numbers multiply to 30 and add to the middle coefficient 11? Let's list all of the factors of 30:
Factors of 30:
1,2,3,5,6,10,15,30
-1,-2,-3,-5,-6,-10,-15,-30 ...List the negative factors as well. This will allow us to find all possible combinations
These factors pair up and multiply to 30
1*30
2*15
3*10
5*6
(-1)*(-30)
(-2)*(-15)
(-3)*(-10)
(-5)*(-6)
note: remember two negative numbers multiplied together make a positive number
Now which of these pairs add to 11? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 11
| First Number | Second Number | Sum | | 1 | 30 | 1+30=31 | | 2 | 15 | 2+15=17 | | 3 | 10 | 3+10=13 | | 5 | 6 | 5+6=11 | | -1 | -30 | -1+(-30)=-31 | | -2 | -15 | -2+(-15)=-17 | | -3 | -10 | -3+(-10)=-13 | | -5 | -6 | -5+(-6)=-11 |
From this list we can see that 5 and 6 add up to 11 and multiply to 30
Now looking at the expression , replace  with  (notice adds up to . So it is equivalent to )
Now let's factor  by grouping:
 Group like terms
 Factor out the GCF of  out of the first group. Factor out the GCF of  out of the second group
 Since we have a common term of  , we can combine like terms
So factors to
So this also means that factors to (since is equivalent to )
------------------------------------------------------------
Answer:
So factors to
Answer by vleith(2983) (Show Source):
You can put this solution on YOUR website! Not quite.
If you multiply your terms, you get 3x^2 -13x -10. Which is not what you want. You are correct in that you need 3&1 and 5&2. Now you need to find the sum of prodcuts that gives you the middle term. Notice that the sign on the constant term (10) is positive. So, the two factor must both have a + or both have a minus, in order to get a positive product.
So, start with this
(3x + A ) (x + B)
The only choices we have that allow the product of A*B to be 10 are 2,5 and 1,10.
We are trying to end up with a sum or products for the middle term that equals 11. Since there is a 3x term in one of the factors, the product can either be 3*10 + 1*1 = 31
OR 3*1 + 1*10 = 13. Neither is 11. So the other pari of factors must be the ones we want. But in which order?
3*2 + 5+1 = 11
3*5 + 2*1 = 17
So the one we want is 3*2 and 5*1
As follows:
(3x+5)(x+2)
Look at this URL --> http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=algebra&s2=factor&s3=basic
|
|
|
