Question 151358: factorize 5(3x+y)^2+6(3x+y)-8 please help me understand how to solve this sum. thankyou
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Let 
So the expression goes from  to 
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 :
1,2,4,5,8,10,20,40
-1,-2,-4,-5,-8,-10,-20,-40
Note: list the negative of each factor. This will allow us to find all possible combinations.
These factors pair up and multiply to .
1*(-40)
2*(-20)
4*(-10)
5*(-8)
(-1)*(40)
(-2)*(20)
(-4)*(10)
(-5)*(8)
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 | -40 | 1+(-40)=-39 | | 2 | -20 | 2+(-20)=-18 | | 4 | -10 | 4+(-10)=-6 | | 5 | -8 | 5+(-8)=-3 | | -1 | 40 | -1+40=39 | | -2 | 20 | -2+20=18 | | -4 | 10 | -4+10=6 | | -5 | 8 | -5+8=3 |
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  from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.
 Combine like terms. Or factor out the common term 
 Plug in
 Distribute
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Answer:
So factors to .
In other words, 
Note: you can check the answer by expanding to get .
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