What are the steps to factor this equation? 6x^2 - 4x - 2?

Use the cross method.

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


Check:


3x+1
2x-2
-------
6x^2+2x
..........-6x-2
-----------------
6x^2-4x-2


Or.


6x^2 - 4x - 2
=2(3x^2-2x-1)

Factor the trinomial.

= (3x+1)(1x-1)

first and foremost, notice each constant can be divided by 2, so factor out a 2:

2(3x^2-2x-1)
then set up 2 factors, knowing that the x term has to multiply to 3x^2, so:
2(3x + )( x + )

then know that the constant terms have to multiply to -1, i.e., 1 and -1 multiply to -1. Then you can see that -1 times 3x plus 1 times x would equal -2x... so:
2(3x + 1)( x - 1)

if you distribute and check your work, you find this is correct!

2 (3x² - 2x - 1)
2 (3x + 1)(x - 1)

6x^2 - 4x - 2 : original equation
= 2(3x^2 - 2x - 1) : take out common factor of 2 to all terms
= 2(3x + 1)(x - 1) : factor inside parentheses by finding two pairs of numbers that multiply to -1 and add to -2. The numbers must be 3, 1 (for the x^2 term) and 1, -1 (for the constant term).

The 3 must multiply to the -1 to get the sum negative, and will have to be the F and L terms (where I placed it), or the O and I terms (if the factors are listed the other way around). The 1's must be the other pair of terms. In other words, the 3 and the -1 are in different parentheses.

6x^2-4x-2=0,x=1 & 1/3,then
2(x-1)(3x+1)

Well, first pull out a 2.

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

Now you need to find two numbers that have a product of -3 and a sum of -2.

Those would be -3, 1

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

You can pull a 3x out of the first binomial.

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

Now group like terms.

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

6x^2 - 4x - 2
= 2(3x^2 - 2x - 1)
= 2(3x^2 + x - 3x - 1)
= 2[(3x^2 + x) - (3x + 1)]
= 2[x(3x + 1) - 1(3x + 1)]
= 2(3x + 1)(x - 1)

6x^2-4x-2
=2(3x^2-2x-1)
=2(3x+1)(x-1)

6x^2 - 4x - 2
6x^2 - 6x + 2x - 2
6x(x - 1) + 2(x -1)
(6x+ 2)(x - 1)

x = -2/6 = -1/3 or x = 1