# Thread: Factor 3x^2 +2x -6

1. ## Factor 3x^2 +2x -6

I'm kind of embarrassed to ask this question since I am in calculus but how would I factor the following? I know you can multiply 3*-6 and you get -18 and somehow you then can factor. But I can't seem to do it.

3x^2 + 2x -6

2. Originally Posted by nautica17
I'm kind of embarrassed to ask this question since I am in calculus but how would I factor the following? I know you can multiply 3*-6 and you get -18 and somehow you then can factor. But I can't seem to do it.

3x^2 + 2x -6

$b^2 - 4ac = 4 - 4(3)(-6) = 4 + 72 = 76$

since the discriminant is not a perfect square number, the quadratic will not factor.

3. Originally Posted by skeeter
$b^2 - 4ac = 4 - 4(3)(-6) = 4 + 72 = 76$

since the discriminant is not a perfect square number, the quadratic will not factor.
Hmm... I see. Well, I'm trying to find the local extrema and that equation happens to be the second derivative. I don't know how to get the max and min point out of that now.

4. Originally Posted by nautica17
Hmm... I see. Well, I'm trying to find the local extrema and that equation happens to be the second derivative. I don't know how to get the max and min point out of that now.
you don't need to know where f''(x) = 0 to find extrema ... inflection points, yes, but not extrema.

if f'(x) = 0 and f''(x) > 0 , you have a min

if f'(x) = 0 and f''(x) < 0 , you have a max

5. Okay, so my f'(x) = x^3 + x^2 -6
From that I got x(x+3)(x-2)

So, increasing at 2 and decreasing at 3, would make my min at 2 and max at 3? Did I get that right?

6. Originally Posted by nautica17
Okay, so my f'(x) = x^3 + x^2 -6
From that I got x(x+3)(x-2)

So, increasing at 2 and decreasing at 3, would make my min at 2 and max at 3? Did I get that right?
check f'(x) again ... did you mean to type f'(x) = x^3 + x^2 - 6x ?

it might be a good idea to see the original function, f(x).

7. Originally Posted by skeeter
check f'(x) again ... did you mean to type f'(x) = x^3 + x^2 - 6x ?

it might be a good idea to see the original function, f(x).

Yea, that's what I meant, sorry.

The original function was ((x^4)/4)+((x^3)/3)-(3x^2)+5

8. Originally Posted by nautica17
Okay, so my f'(x) = x^3 + x^2 -6x
From that I got x(x+3)(x-2)

So, increasing at 2 and decreasing at 3, would make my min at 2 and max at 3? Did I get that right?
sorry, no.

min at x = -3

max at x = 0

min at x = 2

9. Hmm, okay cool, thank-you. I think I get it now.