Factoring a trinomial

lordbodom

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I am working a problem to find the GCF of 3 trinomials. I am having trouble factoring one of them. The trinomial is:

[6x(3)-10x^(2)-4x]

I factored with 2x to get: [2x{3x^(2)-5x-2}]

Now I cant figure out how to factor {3x^(2)-5x-2}. From the other exercies i have practiced, you factor the -2 and see what factorings fit to equal 5. I cant figure out how this fits since for 2, the factors would be 2,1? The solution says (3x+1)(x-2) and doing the multiplication i see that it is correct, but i am not understanding how to get this?
 
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I am working a problem to find the GCF of 3 trinomials. I am having trouble factoring one of them. The trinomial is:

[6x(3)-10x^(2)-4x]

I factored with 2x to get: [2x{3x^(2)-5x-2}]

Now I cant figure out how to factor {3x^(2)-5x-2}. From the other exercies i have practiced, you factor the -2 and see what factorings fit to equal 5. I cant figure out how this fits since for 2, the factors would be 2,1? The solution says > > (3x+1)(x-1) < < and doing the multiplication i see that it is correct, but i am not understanding how to get this?

There's a typo. The factorization of \(\displaystyle \ 3x^2 - 5x - 2 \ \) is (3x + 1)(x - 2).
 
Careful Lord: 6x^3 - 10x^2 - 4x

You got to: 3x^2 - 5x - 2, which is correct...

If difficult to factor, set equal to 0 : 3x^2 - 5x - 2 = 0
Now use the quadratic formula.
Sorry about the typo...using the quadratic formula I get x=-10 and x=-2. Not sure where to go from there?
 
HUH? How did you manage that?

3x^2 - 5x - 2 = 0

x = [-(-5) +- SQRT(5^2 - 4(3)(-2)] / (2(3))

x = [5 +- SQRT(25 + 24)] / 6

x = (5 +- 7) / 6

x = 12/6 or x = -2/6

x = 2 or x = -1/3

Do you FOLLOW that?

I do. Im sorry im a mess, I was reading the WRONG problem in my notepad, was doing a few and for some reason I used the formula on the wrong one. :(
 
the factors for the given expression was (2x)(x-2)(3x+1)

the given expression 6x3 -10x2-4x

take 2x as common then, 2x[3x2-5x-2]

factors for 3x2-5x-2
3*2=6 now find factors for 6, while adding are subtract we should get 5 value
3*2=6 no
6*1=6 yes hence 6-1=5

3x2-6x+1x-2
3x(x-2)+1(x-2)
(x-2)(3x+1)

therefore the factors are (2x)(x-2)(3x+1)
 
the factors for the given expression was (2x)(x-2)(3x+1)

the given expression 6x3 -10x2-4x

take 2x as common then, 2x[3x2-5x-2]

factors for 3x2-5x-2
3*2=6 now find factors for 6, while adding are subtract we should get 5 value
3*2=6 no
6*1=6 yes hence 6-1=5

3x2-6x+1x-2
3x(x-2)+1(x-2)
(x-2)(3x+1)

> > > Therefore, the factors are 2x, x - 2, and 3x+1. < < <

The *factorization* is 2x(x-2)(3x+1).
 
It took me a moment to get it but LookAgain's point is that "the factors are" 2x, x- 2, and 3x+1. 2x(x- 2)(3x+ 1) is "the factorization" or "way of factoring", not just "the factors".
 
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