Question

What are all the possible rational zeros for `f(x)=5x^3+29x^2+19x-5` and how do you find all zeros?

Answer 1

All the rational zeros arefor `x=-1` , `x=-5` and `x=1/5`

Explanation:

By trial and error, we can see that

`f(-1)=-5+29-19-5=0`
So `(x+1)` is a factor
So we do a long division to find the other factors
`5x^3+29x^2+19x-5` `color(white)(aaaaaa)` `∣x+1`
`5x^3+5x^2` `color(white)(aaaaaaaaaaaaaaa)` `∣` `5x^2+24x-5`
`color(white)(a)` `0+24x^2`
`color(white)(aaaaa)` `24x^2+19x`
`color(white)(aaaaaaa)` `0+24x`
`color(white)(aaaaaaaaaaa)` `-5x-5`
`color(white)(aaaaaaaaaaa)` `-5x-5`
`color(white)(aaaaaaaaaaaaaa)` `0+0`
Now we have to factorize
`5x^2+24x-5=(5x-1)(x+5)`

So all the rational zeros arefor `x=-1` , `x=-5` and `x=1/5`