Solve `x^3+x^2-7x+2=0` by first factorising the LHS?

Given: `x^3+x^2-7x+2=0`

`color(blue)("Step 1")`
Consider the constant of 2.

The whole number factors are `1,-1,2,-2`

Test `x=1`

`1^3+1^2-7(1)+2!=0`

Test `x=2`

`2^3+2^2-7(2)+2`

`8+4-14+2=0` so `x=2` is a factor giving:

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

It has to be (-1) as `(-2)xx(-1)=+2`
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`color(blue)("Step 2 - consider the final "x^3" term")`

We require the first term to be `x^3` so the structure has to be:

`(color(red)(x-2))(x^2+.........-1)`

`color(red)(x) xx x^2=x^3` so that is ok!
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`color(blue)("Step 3 - consider the final "x^2" term")`

The negative 2 in the first bracket gives

`color(red)(-2)xx x^2 = -2x^2`

But we need to have `+x^2` so we need to 'build' `+3x^2` to compensate as `+3x^2-2x^2=x^2`

`color(red)(x)color(green)(xx3x)` should do it giving:

`(color(red)(x-2))(x^2color(green)(+3x)+.........-1)`

So what have we got now?

`(color(red)(x-2))(x^2+3x+......-1) -> x^3+3x^2-2x^2-6x -1`
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`color(blue)("Step 4 - consider the final "x" term")`

The target is `-7x` and we have `-6x` so we need another `-x`

So far we have not included the `-1` in the last bracket. lets do so:

Test:

`color(red)((x-2))(x^2+3x-1)`

`color(white)(-)color(red)(x)(x^2+3x-1) color(white)("d")->color(white)("d") x^3+3x^2-x+0`
`color(white)("d")color(red)(-2)(x^2+3x-1)color(white)("d")->ul(color(white)("d.")0-2x^2-6x+2 larr" Add")`
`color(white)("dddddddddddddddddddd")x^3+x^2-7x+2 larr" As required"`
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`color(blue)("Step 5 - Solving for = 0")`

Set `color(red)(x-2)=0 => x=+2`

Set `x^2+3x-1=0`

`x=(-3+-sqrt(3^2-4(1)(-1)))/2`

`x=-3/2+-sqrt(13)/2`

`x~~ +0.3028` to 4 dp
`x~~-3.3028` to 4 dp