How do I factor 2x^2 + 5x - 3

I've gotten up to

x(2x+3) + 1 (2x-3)

now i'm stuck....what do i do if there are two different expressions in the parenthesis?

Try it this way: Look for factors of 2*-3 = -6 that add to 5. These are 6 and -1. So now rewrite your middle term, 5x, as --x + 6x, so you have 2x^2 - x + 6x - 3. Now group each side: (2x^2 - x) + (6x-3) and factor: x(2x-1) + 3(2x-1). Now factor out the common (2x-1) factor: (2x-1)(x+3).

This technique is called "splitting-the-middle," as it splits the middle term of the expression. It helps take some of the trial-and-error out of factoring these types of quadratics. In general, for Ax^2 + Bx + C, you look for factors of A*C that sum to B. Then rewrite the middle term as the sum of those two numbers, group, factor, and factor again. You'll always come up with a product of binomials.

To help reinforce your understanding of these concepts, I've searched and found a webpage and a video tutorial that address problems similar to this one, and I thought they might be helpful to you. I've listed them below.

Use the FOIL method (if you don't know what it is look it up)

You've gotten to the right spot. Set both of the equations equal to zero and then solve for x.

Now you factor 2x + 3, because they're common to both parcels

(2x+3)(x+1)

The parentheses have to contain the same expression to combine.

(2x - 1) (x + 3)

2x^2+5x-3
(2x-1)(x+3)

(2x-1) (x+3)

No, you're doing it wrong.


(2x - 1)(x + 3)