Solve by factoring: 3x^2 + 2x - 1 = 0 {-1, 2/3} {1, -2/3} {-1, 1/3} {1, -1/3}

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Solve by factoring: 3x^2 + 2x - 1 = 0 {-1, 2/3} {1, -2/3} {-1, 1/3} {1, -1/3}

Mathematics
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please explain how i find this
well, dividing first by 3, we have \[ x^2 + \frac{2}{3}x - \frac{1}{3} = 0 \] Now, you want factors of -1/3 that add up to 2/3
@EmmaH, the steps for solving this are the same as what I showed you in the previous problem.

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divide what by three?
I'll even give you a hint, but you have to solve the rest yourself: Given: \(3x^2 + 2x - 1\) Split the middle term: \(3x^2 + 3x - 1x - 1\) Now factor by grouping
(3x^2+3x) + (-1x-1)?
Okay, now what's the next step?
Do you know how to factor \(3x^2 + 3x\) ?
3x(1x+1)
Yes, or simply 3x(x + 1) Writing 1 before the x is not necessary. Now factor (-1x - 1). Let me know what you get.
?
Well, you have a factor of (x+1) above where you wrote 3x(x+1) Can you get a factor of (x+1) out of (-1x - 1) ?
what do you mean?

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