Question

What are the possible rational roots of `2x^3+3x^2-8x+3=0` and then determine the rational roots?

Answer 1

The "possible" rational roots are:

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

The actual roots are:

`1, 1/2, -3`

Explanation:

Given:

`2x^3+3x^2-8x+3=0`

By the rational roots theorem, any rational zeros of this cubic are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `3` and `q` a divisor of the coefficient `2` of the leading term.

That means that the only possible rational zeros are:

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

In addition, note that the sum of the coefficients is `0`. That is:

`2+3-8+3 = 0`

Hence `x=1` is a root and `(x-1)` a factor:

`0 = 2x^3+3x^2-8x+3`

`color(white)(0) = (x-1)(2x^2+5x-3)`

To find the zeros of the remaining quadratic we could try each of the other possible rational roots in turn, but I'd rather complete the square:

`0 = 8(2x^2+5x-3)`

`color(white)(0) = 16x^2+40x-24`

`color(white)(0) = (4x)^2+2(4x)(5)+(5)^2-49`

`color(white)(0) = (4x+5)^2-7^2`

`color(white)(0) = ((4x+5)-7)((4x+5)+7)`

`color(white)(0) = (4x-2)(4x+12)`

`color(white)(0) = (2(2x-1))(4(x+3))`

`color(white)(0) = 8(2x-1)(x+3)`

So the remaining two roots are:

`x=1/2" "` and `" "x=-3`