Question

How do you list all possible roots and find all factors of `6x^3+7x^2-3x-1`?

Answer 1

The "possible" rational zeros are:

`+-1/6, +-1/3, +-1/2, +-1`

The factorisation is:

`6x^3+7x^2-3x-1`

`= 3(2x-1)(x+5/6-sqrt(13)/6)(x+5/6+sqrt(13)/6)`

Explanation:

Given:

`f(x) = 6x^3+7x^2-3x-1`

By the rational roots theorem, any rational zeros of `f(x)` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-1` and `q` a divisor of the coefficient `6` of the leading term.

That means that the only possible rational zeros are:

`+-1/6, +-1/3, +-1/2, +-1`

We find:

`f(1/2) = 6(1/8)+7(1/4)-3(1/2)-1 = (3+7-6-4)/4 = 0`

So `x=1/2` is a zero and `(2x-1)` a factor:

`6x^3+7x^2-3x-1 = (2x-1)(3x^2+5x+1)`

We can factor the remaining quadratic by completing the square.

To reduce the need to do arithmetic with fractions, I will premultiply by `3*2^2 = 12` and divide by it at the end:

`12(3x^2+5x+1) = 36x^2+60x+12`

`color(white)(12(3x^2+5x+1)) = (6x)^2+2(6x)(5)+25-13`

`color(white)(12(3x^2+5x+1)) = (6x+5)^2-(sqrt(13))^2`

`color(white)(12(3x^2+5x+1)) = ((6x+5)-sqrt(13))((6x+5)+sqrt(13))`

`color(white)(12(3x^2+5x+1)) = (6x+5-sqrt(13))(6x+5+sqrt(13))`

`color(white)(12(3x^2+5x+1)) = 12*3(x+5/6-sqrt(13)/6)(x+5/6+sqrt(13)/6)`

So:

`3x^2+60x+12 = 3(x+5/6-sqrt(13)/6)(x+5/6+sqrt(13)/6)`

Putting it all together:

`6x^3+7x^2-3x-1`

`= 3(2x-1)(x+5/6-sqrt(13)/6)(x+5/6+sqrt(13)/6)`