Question

What are all the possible rational zeros for `f(x)=32x^3-52x^2+17x+3` and how do you find all zeros?

Answer 1

`f(x)` has zeros `1`, `-1/8` and `3/4`

Explanation:

`f(x) = 32x^3-52x^2+17x+3`

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 `3` and `q` a divisor of the coefficient `32` of the leading term.

That means that the only possible rational zeros are:

`+-1/32, +-1/16, +-3/32, +-1/8, +-3/16, +-1/4, +-3/8, +-1/2, +-3/4, +-1, +-3/2, +-3`

That is rather a lot of possibilities to try, but note that the sum of the coefficients of `f(x)` is zero. That is:

`32-52+17+3 = 0`

So `f(1) = 0`, `x=1` is a zero and `(x-1)` is a factor:

`32x^3-52x^2+17x+3 = (x-1)(32x^2-20x-3)`

We can find zeros of the remaining quadratic factor using an AC method:

Find a pair of factors of `AC=32*3 = 96` which differ by `B=20`.

The pair `24, 4` works. Use this pair to split the middle term, then factor by grouping:

`32x^2-20x-3`

`=32x^2-24x+4x-3`

`=(32x^2-24x)+(4x-3)`

`=8x(4x-3)+1(4x-3)`

`=(8x+1)(4x-3)`

Hence zeros: `x=-1/8` and `x=3/4`