Yahoo Answers: Answers and Comments for How do I factor 2x^2 + 5x  3 ? [Mathematics]
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From Anonymous
enUS
Tue, 06 Dec 2011 12:24:27 +0000
3
Yahoo Answers: Answers and Comments for How do I factor 2x^2 + 5x  3 ? [Mathematics]
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From Magic Matt: No, you're doing it wrong.
(2x  1)(x...
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Tue, 06 Dec 2011 12:26:55 +0000
No, you're doing it wrong.
(2x  1)(x + 3)

From Crazy Aunt Nell: The parentheses have to contain the same expre...
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Tue, 06 Dec 2011 12:26:13 +0000
The parentheses have to contain the same expression to combine.
(2x  1) (x + 3)

From Keith Roberts: Try it this way: Look for factors of 2*3 = 6...
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Tue, 06 Dec 2011 12:35:48 +0000
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) + (6x3) and factor: x(2x1) + 3(2x1). Now factor out the common (2x1) factor: (2x1)(x+3).
This technique is called "splittingthemiddle," as it splits the middle term of the expression. It helps take some of the trialanderror 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.

From Anonymous: Now you factor 2x + 3, because they're com...
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Tue, 06 Dec 2011 12:30:55 +0000
Now you factor 2x + 3, because they're common to both parcels
(2x+3)(x+1)

From Christina A.: 2x^2+5x3
(2x1)(x+3)
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Tue, 06 Dec 2011 12:27:34 +0000
2x^2+5x3
(2x1)(x+3)

From Michael: Use the FOIL method (if you don't know wha...
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Tue, 06 Dec 2011 12:27:19 +0000
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.

From Andy: (2x1) (x+3)
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Tue, 06 Dec 2011 12:26:30 +0000
(2x1) (x+3)