How do you factor 2x^2-5x-3=0?
I have the answer. I know that it factors out like this:
(2x+1)(x-3)=0 and then continues to get the answer x=-1/2 or x=3. i get everything but the part in the parentheses so finishing the entire problem is not necessary.
The part I don't understand is why the 2x is in the first parentheses instead of the second. How do you tell which one to put it in?
Also, why doesn't it add up to -5? ...
any help would be greatly appreciated!
Comments
The 2x is in the first factor because if you put it in the second factor and then multiply out (x+1)(2x-3) (using FOIL or any other correct method), you don't get 2x^2-5x-3. You get 2x^2 - x - 3.
In your second question, the solutions would only add up to -5 if the first term were x^2, instead of 2x^2. The 'rule' about the solutions adding up to the middle coefficient only applies of the x^2 term has a coefficient of 1.
you know that the factors can have the form [2x -a][x-b] , since the quadratic starts with 2 x²....now you also need : ab = -3{ thus a & b have opposite signs} and -2b-a = -5......if you are looking for integers then the only pairs are [1,-3] , [-1,3]....obviously the last with a = -1 & b = 3
(2x+1)*x = 2x^2 + x (expression 1)
(2x+1)*-3 = -6x - 3 (expression 2)
Now sum expression 1 and expression 2:
2x^2 + x - 6x - 3 = 2x^2 -5x - 3
Does it add up now?
The answer is -10