SOLUTION: How do you factor 3x^2+x-2 ?

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Question 888565: How do you factor 3x^2+x-2 ?
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)


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



Now multiply the first coefficient 3 by the last term -2 to get %283%29%28-2%29=-6.



Now the question is: what two whole numbers multiply to -6 (the previous product) and add to the second coefficient 1?



To find these two numbers, we need to list all of the factors of -6 (the previous product).



Factors of -6:

1,2,3,6

-1,-2,-3,-6



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



These factors pair up and multiply to -6.

1*(-6) = -6
2*(-3) = -6
(-1)*(6) = -6
(-2)*(3) = -6


Now let's add up each pair of factors to see if one pair adds to the middle coefficient 1:



First NumberSecond NumberSum
1-61+(-6)=-5
2-32+(-3)=-1
-16-1+6=5
-23-2+3=1




From the table, we can see that the two numbers -2 and 3 add to 1 (the middle coefficient).



So the two numbers -2 and 3 both multiply to -6 and add to 1



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



3x%5E2%2Bhighlight%28-2x%2B3x%29-2 Replace the second term 1x with -2x%2B3x.



%283x%5E2-2x%29%2B%283x-2%29 Group the terms into two pairs.



x%283x-2%29%2B%283x-2%29 Factor out the GCF x from the first group.



x%283x-2%29%2B1%283x-2%29 Factor out 1 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%2B1%29%283x-2%29 Combine like terms. Or factor out the common term 3x-2



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



Answer:



So 3%2Ax%5E2%2Bx-2 factors to %28x%2B1%29%283x-2%29.



In other words, 3%2Ax%5E2%2Bx-2=%28x%2B1%29%283x-2%29.



Note: you can check the answer by expanding %28x%2B1%29%283x-2%29 to get 3%2Ax%5E2%2Bx-2 or by graphing the original expression and the answer (the two graphs should be identical).


Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 3x%5E2%2B1x%2B-2+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%281%29%5E2-4%2A3%2A-2=25.

Discriminant d=25 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-1%2B-sqrt%28+25+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%281%29%2Bsqrt%28+25+%29%29%2F2%5C3+=+0.666666666666667
x%5B2%5D+=+%28-%281%29-sqrt%28+25+%29%29%2F2%5C3+=+-1

Quadratic expression 3x%5E2%2B1x%2B-2 can be factored:
3x%5E2%2B1x%2B-2+=+3%28x-0.666666666666667%29%2A%28x--1%29
Again, the answer is: 0.666666666666667, -1. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+3%2Ax%5E2%2B1%2Ax%2B-2+%29