SOLUTION: 15y^2-y-2

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Question 599416: 15y^2-y-2
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Looking at the expression 15y%5E2-y-2, we can see that the first coefficient is 15, the second coefficient is -1, and the last term is -2.


Now multiply the first coefficient 15 by the last term -2 to get %2815%29%28-2%29=-30.


Now the question is: what two whole numbers multiply to -30 (the previous product) and add to the second coefficient -1?


To find these two numbers, we need to list all of the factors of -30 (the previous product).


Factors of -30:
1,2,3,5,6,10,15,30
-1,-2,-3,-5,-6,-10,-15,-30


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


These factors pair up and multiply to -30.
1*(-30) = -30
2*(-15) = -30
3*(-10) = -30
5*(-6) = -30
(-1)*(30) = -30
(-2)*(15) = -30
(-3)*(10) = -30
(-5)*(6) = -30

Now let's add up each pair of factors to see if one pair adds to the middle coefficient -1:


First NumberSecond NumberSum
1-301+(-30)=-29
2-152+(-15)=-13
3-103+(-10)=-7
5-65+(-6)=-1
-130-1+30=29
-215-2+15=13
-310-3+10=7
-56-5+6=1



From the table, we can see that the two numbers 5 and -6 add to -1 (the middle coefficient).


So the two numbers 5 and -6 both multiply to -30 and add to -1


Now replace the middle term -1y with 5y-6y. Remember, 5 and -6 add to -1. So this shows us that 5y-6y=-1y.


15y%5E2%2Bhighlight%285y-6y%29-2 Replace the second term -1y with 5y-6y.


%2815y%5E2%2B5y%29%2B%28-6y-2%29 Group the terms into two pairs.


5y%283y%2B1%29%2B%28-6y-2%29 Factor out the GCF 5y from the first group.


5y%283y%2B1%29-2%283y%2B1%29 Factor out 2 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.


%285y-2%29%283y%2B1%29 Combine like terms. Or factor out the common term 3y%2B1


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Answer:


So 15y%5E2-y-2 factors to %285y-2%29%283y%2B1%29.


In other words, 15y%5E2-y-2=%285y-2%29%283y%2B1%29.


Note: you can check the answer by expanding %285y-2%29%283y%2B1%29 to get 15y%5E2-y-2 or by graphing the original expression and the answer (the two graphs should be identical).