Question

`4x^3+2x^2-2x-1color(white)(.)`?

Answer 1

I'm assuming you want this to be factored. We have:

`=2x^2(2x + 1) - (2x + 1)`

`=(2x^2 - 1)(2x + 1)`

This is as far as we can go as `2x+ 1` is in simplest form and `2x^2 - 1` cannot be factored as a difference of squares with rational coefficients.

Hopefully this helps!

Answer 2

`x=-1/2 and x=sqrt2/2 and x=-sqrt2/2`

Explanation:

Lets try 'playing' with the numbers so that they look different but are overall the same in value.

Set `color(blue)(4x^3+2x^2)color(green)(-2x-1)=y=0`

We need to try and 'force' the numbers to give something we can work with.

Split it like this:

Note that `color(magenta)(-(2xcolor(white)("dd")+color(white)("ddd")1))` is the same as
`[color(white)(d/d)(-1)xx2xcolor(white)("d")]+[color(white)(d/d)(-1)xx(+1)color(white)("d")] = color(green)(-2x-1)`

Split it like this:

`color(blue)((4x^3+2x^2))color(magenta)(-(2x+1))=y=0`

Note that 1 times anything does not change it. So we can write
`-(2x+1)" as "-1(2x+1)`

`color(blue)((4x^3+2x^2))color(magenta)(-1(2x+1))=y=0`

Factor out `2x^2` from the `(4x^3+2x^2)`

`color(blue)(2x^2(2x+1)) color(magenta)(-1(2x+1))=y=0`

Now we can factor out the `(2x+1)` giving:

`(2x+1)(2x^2-1)=0`

The temptation is to stop hear but that is a mistake.

Consider the `(2x^2-1)` can you spot something?

This is the same as `(2x^2-1^2) =(sqrt(2)color(white)("d")x-1)(sqrt(2)color(white)("d")x+1)`

Or to make it more obvious:
`(2x^2-1^2) =(xsqrt(2)-1)(xsqrt(2)+1)`

So putting it all together we have:

`4x^3+2x^2-2x-1=y=0=(2x+1)(x sqrt(2)-1)(x sqrt(2)+1)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
For each of these to be 0 we have:

`2x+1=0color(white)("d") =>color(white)("d") 2x=-1color(white)("d") =>color(white)("d") color(red)(x=-1/2)`

`xsqrt(2)-1=0color(white)("d")=>color(white)("d") xsqrt(2)=1color(white)("d")=>color(white)("d")color(red)(x=1/sqrt(2) = sqrt2/2)`

`xsqrt(2)+1=0color(white)("d")=>color(white)("d")xsqrt(2)=-1color(white)("d")=>color(white)("d")color(red)(x=-1/sqrt(2) = -sqrt2/2)`