SOLUTION: I know how to factorise equations and solve such as x^2+4x-21=0 already, but Im not sure how to do it when there is a number in front of x^2. For example: 3b^2+5b+2=0 How could I

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: I know how to factorise equations and solve such as x^2+4x-21=0 already, but Im not sure how to do it when there is a number in front of x^2. For example: 3b^2+5b+2=0 How could I      Log On


   

Question 659398: I know how to factorise equations and solve such as x^2+4x-21=0 already, but Im not sure how to do it when there is a number in front of x^2. For example:
3b^2+5b+2=0
How could I factorize and then solve this equation?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Looking at the expression 3b%5E2%2B5b%2B2, we can see that the first coefficient is 3, the second coefficient is 5, and the last term is 2.


Now multiply the first coefficient 3 by the last term 2 to get %283%29%282%29=6.


Now the question is: what two whole numbers multiply to 6 (the previous product) and add to the second coefficient 5?


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 5:


First NumberSecond NumberSum
161+6=7
232+3=5
-1-6-1+(-6)=-7
-2-3-2+(-3)=-5



From the table, we can see that the two numbers 2 and 3 add to 5 (the middle coefficient).


So the two numbers 2 and 3 both multiply to 6 and add to 5


Now replace the middle term 5b with 2b%2B3b. Remember, 2 and 3 add to 5. So this shows us that 2b%2B3b=5b.


3b%5E2%2Bhighlight%282b%2B3b%29%2B2 Replace the second term 5b with 2b%2B3b.


%283b%5E2%2B2b%29%2B%283b%2B2%29 Group the terms into two pairs.


b%283b%2B2%29%2B%283b%2B2%29 Factor out the GCF b from the first group.


b%283b%2B2%29%2B1%283b%2B2%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.


%28b%2B1%29%283b%2B2%29 Combine like terms. Or factor out the common term 3b%2B2



So 3b%5E2%2B5b%2B2 factors to %28b%2B1%29%283b%2B2%29.


In other words, 3b%5E2%2B5b%2B2=%28b%2B1%29%283b%2B2%29.

I'll leave the rest to you.

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