SOLUTION: simplify 2n^2+7n+5

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: simplify 2n^2+7n+5      Log On


   

Question 303063: simplify 2n^2+7n+5
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Looking at the expression 2n%5E2%2B7n%2B5, we can see that the first coefficient is 2, the second coefficient is 7, and the last term is 5.


Now multiply the first coefficient 2 by the last term 5 to get %282%29%285%29=10.


Now the question is: what two whole numbers multiply to 10 (the previous product) and add to the second coefficient 7?


To find these two numbers, we need to list all of the factors of 10 (the previous product).


Factors of 10:
1,2,5,10
-1,-2,-5,-10


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


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

Now let's add up each pair of factors to see if one pair adds to the middle coefficient 7:


First NumberSecond NumberSum
1101+10=11
252+5=7
-1-10-1+(-10)=-11
-2-5-2+(-5)=-7



From the table, we can see that the two numbers 2 and 5 add to 7 (the middle coefficient).


So the two numbers 2 and 5 both multiply to 10 and add to 7


Now replace the middle term 7n with 2n%2B5n. Remember, 2 and 5 add to 7. So this shows us that 2n%2B5n=7n.


2n%5E2%2Bhighlight%282n%2B5n%29%2B5 Replace the second term 7n with 2n%2B5n.


%282n%5E2%2B2n%29%2B%285n%2B5%29 Group the terms into two pairs.


2n%28n%2B1%29%2B%285n%2B5%29 Factor out the GCF 2n from the first group.


2n%28n%2B1%29%2B5%28n%2B1%29 Factor out 5 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.


%282n%2B5%29%28n%2B1%29 Combine like terms. Or factor out the common term n%2B1


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


Answer:


So 2n%5E2%2B7n%2B5 factors to %282n%2B5%29%28n%2B1%29.


In other words, 2n%5E2%2B7n%2B5=%282n%2B5%29%28n%2B1%29.


Note: you can check the answer by expanding %282n%2B5%29%28n%2B1%29 to get 2n%5E2%2B7n%2B5 or by graphing the original expression and the answer (the two graphs should be identical).