Question

What are all the possible rational zeros for `f(x)=5x^3+x^2-5x-1` and how do you find all zeros?

Answer 1

The possible rational zeros are `-1, -1/5, 1/5, 5` and the zeros are `-1, -1/5, 1`

Explanation:

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

To find all the possible rational zeros, divide all the factors `p` of the constant term by all the factors `q` of the leading coefficient. The list of possible rational zeros is given by `p/q`.

The constant term `=color(blue)1` and the leading coefficient `=color(red)5`.

The factors `p` of the constant term `color(blue)1` are `+-1`.
The factors `q` of the leading coefficient `color(red)5` are `+-1` and `+-5`

`p/q`=`frac{+-1}{+-1,+-5}=1, -1, 1/5, -1/5`

The possible rational zeros are `-1, -1/5,1/5,1`

To find all the zeros, factor the polynomial and set each factor equal to zero.

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

Factor by grouping.
`(5x^3+x^2)-(5x+1)`
`x^2(5x+1)-1(5x+1)`
`(x^2-1)(5x+1)`

Use the difference of squares to factor `x^2-1`
`(x+1)(x-1)(5x+1)`

Set each factor equal to zero and solve.
`x+1=0color(white)(aaa)x-1=0color(white)(aaa)5x+1=0`

`x=-1color(white)(aaa)x=1color(white)(aaa)x=-1/5`