by spoke
 P: 5 2x^2+9x-5 here is the quadratic expression that im trying to factor. Is there a way to factor this easily? im getting confused on the signs. the book says the factor are: (2x-1)(x+5) but (-1)(2)+5=3 but why not (2x+5)(x-1)? and 5(2)=10-1=9 which looks like the correct answer to me.
 P: 5 You wrote "(2x + A)(x + B)" and "when adding A + 2B, they must give 3." for the equation 2x^2 + 3x - 5 Now my question is why 2B + A and not 2A + B? it is always like this? im wondering because the coefficient 2 is in the same parenthesis as A in (2x + A)(x + B). So if it was a 3 instead would it be 3A + B? Im just really confused with how you solved this. why not just use the quadratic formula? wouldnt it be easier and faster? Edit: I just noticed when i plugged the example equation you gave of 2x^2 + 3x -5 = 0 into the quadratic formula i got the solutions (x=1), (x=-2.5) why is this different from your solutions?
P: 4,546

 Quote by spoke You wrote "(2x + A)(x + B)" and "when adding A + 2B, they must give 3." for the equation 2x^2 + 3x - 5 Now my question is why 2B + A and not 2A + B? it is always like this? im wondering because the coefficient 2 is in the same parenthesis as A in (2x + A)(x + B). So if it was a 3 instead would it be 3A + B? Im just really confused with how you solved this. why not just use the quadratic formula? wouldnt it be easier and faster? Edit: I just noticed when i plugged the example equation you gave of 2x^2 + 3x -5 = 0 into the quadratic formula i got the solutions (x=1), (x=-2.5) why is this different from your solutions?
Hey spoke.

Finding the roots of the equation (2x^2 + 3x - 5 = 0, find x from this) is not the same as factoring.

For the factorization 2x^2 + 3x - 5 = (2x + A)(x + B), the roots correspond to the situations when each factor is zero. In other words (x=1) and (x=-2.5) correspond to the situations when (2x + A) = 0 for one of these x's and when (x+B) = 0 for the other x value (i.e. the values x=1 and x=-2.5).

In terms of the factorization this might make it a little clearer:

(2x + A)(x + B) = 2x^2 + 2Bx + Ax + AB = 2x^2 + x(2B+A) + AB

Now we have to match up 2x^2 + 3x - 5 with 2x^2 + x(2B+A) + AB. In other words.

2x^2 + 3x - 5 = 2x^2 + (2B+A)x + AB

This means 2B+A = 2 and AB = -5. We then solve for A and B to get our factorization.
 P: 687 Hi, spoke, let me add something: To factor a quadratic, you must first be fluent in expanding parentheses; that was my point. You ask, why 2B + A and not 2A + B? Try expanding (2x + A)(x + B) into the form Something.x^2 + Something.x + Something, and you'll find out.
P: 88
 Quote by spoke . . . the book says the factor are: (2x-1)(x+5) . . .
The book is correct. If you multiply this expression out, you get back the original equation: 2x^2+9x-5

Another option you may consider: use the Quadratic Formula. (Have you covered the Quadratic Formula yet?) If I find myself fiddling around with an equation too long, sometimes it is easier to go right to the Quadratic Formula. Then you get the two roots--even for situations in which the answers are not nice round numbers. And once you have them, you then know the equation can be put into the form (x - root1)(x - root2).
 P: 2 2x^2+9x-5=2x^2+10x-x-5 =2x(x+5)-(x+5) =(x+5)(2x-1)

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