Question

What are all the possible rational zeros for `f(x)=x^3+x^2-19x+5` and how do you find all zeros?

Answer 1

The possible rational zeros are `1, -1, 5, -5`.
The zeros are `x=-5, x=2+sqrt3, x=2-sqrt3`.

Explanation:

`f(x)=color(blue)1x^3+x^2-19x+color(red)5`

The possible rational zeros are the factors of the constant term `color(red)5` divided by the factors of the leading coefficient `color(blue)1`.

`x=(+-1,+-5)/(+-1) =1, -1,5, -5`

To find all the zeros, choose one of the possible zeros and use synthetic division. If the remainder is zero, you have verified that the possible zero is in fact a zero.

Choosing `-5` as a possible zero:

`color(white)---------`
`-5]color(white)a1color(white)(aaaa)1color(white)(aa)-19color(white)(ax^111)5`
`color(white)(a^1a^1)darrcolor(white)(1)-5color(white)(a^1+)20color(white)(x^1)-5`
`color(white)---------`
`color(white)(a^1a^11)1color(white)(x^2)-4color(white)(aaaaa)1color(white)(aax^11)0`

The remainder is zero, so `-5` is confirmed as a zero.

The coefficients of the result of synthetic division give
`x^2-4x+1`. Because this is not factorable, use the quadratic formula to find the last two zeros.

`x=(4+-sqrt((-4)^2-4*1*1))/2=(4+-sqrt12)/2=(4+-2sqrt3)/2`

`x=2+-sqrt3`

The zeros are `x=-5, x=2+sqrt3, x=2-sqrt3`