Question

How do you factor `8x^2 - 16x + 6 = 0`?

Answer 1

`x = 1/2, 3/2`

Explanation:

We have,

`color(white)(xxx)8x^2 - 16x + 6 = 0`

`rArr 2(4x^2 - 8x + 3) = 0` [Take 2 as common]

`rArr 4x^2 - 8x + 3 = 0` [Divide both sides by `2`]

`rArr 4x^2 - (2 + 6)x + 3 = 0` [Break `8` as `2 + 6`]

`rArr 4x^2 - 2x - 6x + 3 = 0` [Use Distributive Property]

`rArr 2x(2x - 1) - 3(2x - 1) = 0` [Group the like terms]

`rArr (2x - 1)(2x - 3) = 0` [Again Group the like terms]

Now, We know, when the product of two real numbers is zero, then one of them or both of them are zero.

So,

If `2x - 1 = 0 rArr 2x = 1 rArr x = 1/2`

And If `2x - 3 = 0 rArr 2x = 3 rArr x = 3/2`

So, The solution is `x = 1/2, 3/2`.

Hope this helps.

Answer 2

`(2x-3)(4x-2)`

Try to focus on the factors of `8` and the factors of `6`. The `-16` will come if you keep trying. Also, keep in mind that it is a negative `16`.

Explanation:

I solved this by thinking to myself...

Okay, what are some factors of `8`?

`2 xx 4`

`1 xx 8`

Usually, the outermost numbers don't seem to work so I tried `2` and `4` on my first try. I knew that the `8` had to be positive so both of these numbers had to be negative or they both had to be positive.

Then, I thought about factors of six, knowing that the six must be positive.

`1 xx 6`

`2 xx 3`

(or the negative versions of these)

But since I knew that the middle coefficient was `-16`, I figured I would try the negative numbers first.

`(2x - 3) (4x - 2)`

I chose this order of the numbers because I knew that it would be difficult to get a sixteen, since it is considerably larger than these numbers. This is why I chose the `3` to be in the opposite set of numbers than the `4`.

Knowing that I could get `-12` from this set. With time you will become better at factoring.