SOLUTION: Factor this trinomial: 2t^2 - t - 3

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Question 694628: Factor this trinomial:
2t^2 - t - 3
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Looking at the expression 2t%5E2-t-3, we can see that the first coefficient is 2, the second coefficient is -1, and the last term is -3.


Now multiply the first coefficient 2 by the last term -3 to get %282%29%28-3%29=-6.


Now the question is: what two whole numbers multiply to -6 (the previous product) and add to the second coefficient -1?


To find these two numbers, we need to list all of the factors of -6 (the previous product).


Factors of -6:
1,2,3,6
-1,-2,-3,-6


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


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

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


First NumberSecond NumberSum
1-61+(-6)=-5
2-32+(-3)=-1
-16-1+6=5
-23-2+3=1



From the table, we can see that the two numbers 2 and -3 add to -1 (the middle coefficient).


So the two numbers 2 and -3 both multiply to -6 and add to -1


Now replace the middle term -1t with 2t-3t. Remember, 2 and -3 add to -1. So this shows us that 2t-3t=-1t.


2t%5E2%2Bhighlight%282t-3t%29-3 Replace the second term -1t with 2t-3t.


%282t%5E2%2B2t%29%2B%28-3t-3%29 Group the terms into two pairs.


2t%28t%2B1%29%2B%28-3t-3%29 Factor out the GCF 2t from the first group.


2t%28t%2B1%29-3%28t%2B1%29 Factor out 3 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.


%282t-3%29%28t%2B1%29 Combine like terms. Or factor out the common term t%2B1


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Answer:


So 2t%5E2-t-3 factors to %282t-3%29%28t%2B1%29.


In other words, 2t%5E2-t-3=%282t-3%29%28t%2B1%29.


Note: you can check the answer by expanding %282t-3%29%28t%2B1%29 to get 2t%5E2-t-3 or by graphing the original expression and the answer (the two graphs should be identical).