SOLUTION: 5x^2+21x+4

Algebra ->  Equations -> SOLUTION: 5x^2+21x+4      Log On


   



Question 741455: 5x^2+21x+4
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Looking at the expression 5x%5E2%2B21x%2B4, we can see that the first coefficient is 5, the second coefficient is 21, and the last term is 4.


Now multiply the first coefficient 5 by the last term 4 to get %285%29%284%29=20.


Now the question is: what two whole numbers multiply to 20 (the previous product) and add to the second coefficient 21?


To find these two numbers, we need to list all of the factors of 20 (the previous product).


Factors of 20:
1,2,4,5,10,20
-1,-2,-4,-5,-10,-20


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


These factors pair up and multiply to 20.
1*20 = 20
2*10 = 20
4*5 = 20
(-1)*(-20) = 20
(-2)*(-10) = 20
(-4)*(-5) = 20

Now let's add up each pair of factors to see if one pair adds to the middle coefficient 21:


First NumberSecond NumberSum
1201+20=21
2102+10=12
454+5=9
-1-20-1+(-20)=-21
-2-10-2+(-10)=-12
-4-5-4+(-5)=-9



From the table, we can see that the two numbers 1 and 20 add to 21 (the middle coefficient).


So the two numbers 1 and 20 both multiply to 20 and add to 21


Now replace the middle term 21x with x%2B20x. Remember, 1 and 20 add to 21. So this shows us that x%2B20x=21x.


5x%5E2%2Bhighlight%28x%2B20x%29%2B4 Replace the second term 21x with x%2B20x.


%285x%5E2%2Bx%29%2B%2820x%2B4%29 Group the terms into two pairs.


x%285x%2B1%29%2B%2820x%2B4%29 Factor out the GCF x from the first group.


x%285x%2B1%29%2B4%285x%2B1%29 Factor out 4 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.


%28x%2B4%29%285x%2B1%29 Combine like terms. Or factor out the common term 5x%2B1


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


Answer:


So 5x%5E2%2B21x%2B4 factors to %28x%2B4%29%285x%2B1%29.


In other words, 5x%5E2%2B21x%2B4=%28x%2B4%29%285x%2B1%29.


Note: you can check the answer by expanding %28x%2B4%29%285x%2B1%29 to get 5x%5E2%2B21x%2B4 or by graphing the original expression and the answer (the two graphs should be identical).