SOLUTION: factor trinomial using trial and error 6x^2+39x+60 or is it not factorable?

Algebra ->  Equations -> SOLUTION: factor trinomial using trial and error 6x^2+39x+60 or is it not factorable?       Log On


   

Question 521778: factor trinomial using trial and error
6x^2+39x+60
or is it not factorable?

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

6x%5E2%2B39x%2B60 Start with the given expression.


3%282x%5E2%2B13x%2B20%29 Factor out the GCF 3.


Now let's try to factor the inner expression 2x%5E2%2B13x%2B20


---------------------------------------------------------------


Looking at the expression 2x%5E2%2B13x%2B20, we can see that the first coefficient is 2, the second coefficient is 13, and the last term is 20.


Now multiply the first coefficient 2 by the last term 20 to get %282%29%2820%29=40.


Now the question is: what two whole numbers multiply to 40 (the previous product) and add to the second coefficient 13?


To find these two numbers, we need to list all of the factors of 40 (the previous product).


Factors of 40:
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 40.
1*40 = 40
2*20 = 40
4*10 = 40
5*8 = 40
(-1)*(-40) = 40
(-2)*(-20) = 40
(-4)*(-10) = 40
(-5)*(-8) = 40

Now let's add up each pair of factors to see if one pair adds to the middle coefficient 13:


First NumberSecond NumberSum
1401+40=41
2202+20=22
4104+10=14
585+8=13
-1-40-1+(-40)=-41
-2-20-2+(-20)=-22
-4-10-4+(-10)=-14
-5-8-5+(-8)=-13



From the table, we can see that the two numbers 5 and 8 add to 13 (the middle coefficient).


So the two numbers 5 and 8 both multiply to 40 and add to 13


Now replace the middle term 13x with 5x%2B8x. Remember, 5 and 8 add to 13. So this shows us that 5x%2B8x=13x.


2x%5E2%2Bhighlight%285x%2B8x%29%2B20 Replace the second term 13x with 5x%2B8x.


%282x%5E2%2B5x%29%2B%288x%2B20%29 Group the terms into two pairs.


x%282x%2B5%29%2B%288x%2B20%29 Factor out the GCF x from the first group.


x%282x%2B5%29%2B4%282x%2B5%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%282x%2B5%29 Combine like terms. Or factor out the common term 2x%2B5


--------------------------------------------------


So 3%282x%5E2%2B13x%2B20%29 then factors further to 3%28x%2B4%29%282x%2B5%29


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


Answer:


So 6x%5E2%2B39x%2B60 completely factors to 3%28x%2B4%29%282x%2B5%29.


In other words, 6x%5E2%2B39x%2B60=3%28x%2B4%29%282x%2B5%29.


Note: you can check the answer by expanding 3%28x%2B4%29%282x%2B5%29 to get 6x%5E2%2B39x%2B60 or by graphing the original expression and the answer (the two graphs should be identical).

If you need more help, email me at jim_thompson5910@hotmail.com

Also, please consider visiting my website: http://www.freewebs.com/jimthompson5910/home.html and making a donation. Thank you

Jim