SOLUTION: Please help me solve this problem 6c^2+13c+6

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Question 157262This question is from textbook algebra 2
: Please help me solve this problem 6c^2+13c+6 This question is from textbook algebra 2

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
I'm assuming that you want to factor.


Looking at the expression 6c%5E2%2B13c%2B6, we can see that the first coefficient is 6, the second coefficient is 13, and the last term is 6.


Now multiply the first coefficient 6 by the last term 6 to get %286%29%286%29=36.


Now the question is: what two whole numbers multiply to 36 (the previous product) and add to the second coefficient 13?


To find these two numbers, we need to list all of the factors of 36 (the previous product).


Factors of 36:
1,2,3,4,6,9,12,18,36
-1,-2,-3,-4,-6,-9,-12,-18,-36


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


These factors pair up and multiply to 36.
1*36
2*18
3*12
4*9
6*6
(-1)*(-36)
(-2)*(-18)
(-3)*(-12)
(-4)*(-9)
(-6)*(-6)

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


First NumberSecond NumberSum
1361+36=37
2182+18=20
3123+12=15
494+9=13
666+6=12
-1-36-1+(-36)=-37
-2-18-2+(-18)=-20
-3-12-3+(-12)=-15
-4-9-4+(-9)=-13
-6-6-6+(-6)=-12



From the table, we can see that the two numbers 4 and 9 add to 13 (the middle coefficient).


So the two numbers 4 and 9 both multiply to 36 and add to 13


Now replace the middle term 13c with 4c%2B9c. Remember, 4 and 9 add to 13. So this shows us that 4c%2B9c=13c.


6c%5E2%2Bhighlight%284c%2B9c%29%2B6 Replace the second term 13c with 4c%2B9c.


%286c%5E2%2B4c%29%2B%289c%2B6%29 Group the terms into two pairs.


2c%283c%2B2%29%2B%289c%2B6%29 Factor out the GCF 2c from the first group.


2c%283c%2B2%29%2B3%283c%2B2%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.


%282c%2B3%29%283c%2B2%29 Combine like terms. Or factor out the common term 3c%2B2

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Answer:


So 6c%5E2%2B13c%2B6 factors to %282c%2B3%29%283c%2B2%29.


Note: you can check the answer by FOILing %282c%2B3%29%283c%2B2%29 to get 6c%5E2%2B13c%2B6 or by graphing the original expression and the answer (the two graphs should be identical).


Questions? Email me at jim_thompson5910@hotmail.com