Question

What are all the possible rational zeros for `f(x)=5x^3-11x^2+7x-1` and how do you find all zeros?

Answer 1

Zeros are `x=0.2 and x=1`

Explanation:

`f(x) = 5x^3 -11x^2 +7x-1 = 5x^3 -5x^2-6x^2+6x +x-1`

`= 5x^2 (x-1)- 6x (x-1) +1 (x-1)`

`= (x-1)(5x^2-6x+1) = (x-1)(5x^2-5x-x+1)`

`= (x-1){5x(x-1) -1(x-1} = (x-1)(x-1)( 5x-1)`

`= (x-1)^2(5x-1)`

Zeros are

`(x-1=0 or x=1 and 5x-1=0 or 5x=1 or x=0.2`

graph{5x^3-11x^2+7x-1 [-10, 10, -5, 5]}
[Ans]

Answer 2

The "possible" rational zeros are:

`+-1/5, +-1`

The actual zeros are:

`1" "` with multiplicity `2`

`1/5" "` with multiplicity `1`

Explanation:

Given:

`f(x) = 5x^3-11x^2+7x-1`

By the rational root 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 `-1` and `q` a divisor of the coefficient `5` of the leading term.

That means that the only possible rational zeros are:

`+-1/5, +-1`

To find the actual zeros, we could just try each of these in turn, but there is a shortcut:

Note that the sum of the coefficients of `f(x)` is zero. That is:

`5-11+7-1 = 0`

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

`5x^3-11x^2+7x-1 = (x-1)(5x^2-6x+1)`

The same is true of the remaining quadratic:

`5-6+1 = 0`

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

`5x^2-6x+1 = (x-1)(5x-1)`

The remaining linear factor `5x-1` gives us a zero `x=1/5`

So the zeros of `f(x)` are:

`1" "` with multiplicity `2`

`1/5" "` with multiplicity `1`

graph{5x^3-11x^2+7x-1 [-2.107, 2.89, -1.14, 1.36]}