SOLUTION: How do you factor Quadratic Equations? Please help, I have an exam tomorrow and I am so lost. I don't understand a thing...

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Question 122298: How do you factor Quadratic Equations? Please help, I have an exam tomorrow and I am so lost. I don't understand a thing...
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Let's say we want to factor 3x%5E2%2B12x%2B9


Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)


3%2Ax%5E2%2B12%2Ax%2B9 Start with the given expression.



3%28x%5E2%2B4x%2B3%29 Factor out the GCF 3.



Now let's try to factor the inner expression x%5E2%2B4x%2B3



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Looking at the expression x%5E2%2B4x%2B3, we can see that the first coefficient is 1, the second coefficient is 4, and the last term is 3.



Now multiply the first coefficient 1 by the last term 3 to get %281%29%283%29=3.



Now the question is: what two whole numbers multiply to 3 (the previous product) and add to the second coefficient 4?



To find these two numbers, we need to list all of the factors of 3 (the previous product).



Factors of 3:

1,3

-1,-3



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



These factors pair up and multiply to 3.

1*3 = 3
(-1)*(-3) = 3


Now let's add up each pair of factors to see if one pair adds to the middle coefficient 4:



First NumberSecond NumberSum
131+3=4
-1-3-1+(-3)=-4




From the table, we can see that the two numbers 1 and 3 add to 4 (the middle coefficient).



So the two numbers 1 and 3 both multiply to 3 and add to 4



Now replace the middle term 4x with x%2B3x. Remember, 1 and 3 add to 4. So this shows us that x%2B3x=4x.



x%5E2%2Bhighlight%28x%2B3x%29%2B3 Replace the second term 4x with x%2B3x.



%28x%5E2%2Bx%29%2B%283x%2B3%29 Group the terms into two pairs.



x%28x%2B1%29%2B%283x%2B3%29 Factor out the GCF x from the first group.



x%28x%2B1%29%2B3%28x%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.



%28x%2B3%29%28x%2B1%29 Combine like terms. Or factor out the common term x%2B1



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So 3%28x%5E2%2B4x%2B3%29 then factors further to 3%28x%2B3%29%28x%2B1%29



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Answer:



So 3%2Ax%5E2%2B12%2Ax%2B9 completely factors to 3%28x%2B3%29%28x%2B1%29.



In other words, 3%2Ax%5E2%2B12%2Ax%2B9=3%28x%2B3%29%28x%2B1%29.



Note: you can check the answer by expanding 3%28x%2B3%29%28x%2B1%29 to get 3%2Ax%5E2%2B12%2Ax%2B9 or by graphing the original expression and the answer (the two graphs should be identical).