Question

What are all the possible rational zeros for `f(x)=x^3+5x^2-48x-10` and how do you find all zeros?

Answer 1

The only rational zero is `5`.

Explanation:

As putting `x=5` in `f(x)=x^3+5x^2-48x-10`

gives `f(x)=5^3+5xx5^2-48xx5-10=125+125-240-10=0`

Hence, `5` is a zero of `f(x)` and `(x-5)` is a factor of `f(x)`

Hence `f(x)=x^3+5x^2-48x-10`

= `x^2(x-5)+10x(x-5)+2(x-5)`

= `(x-5)(x^2+10x+2)`

As the determinant in `x^2+10x+2` is `10^2-4xx1xx2=100-8=92`, which though positive, is not a square of a rational number, there are no other rational zeros.

Irrational zeros can be had using quadratic formula as `(-10+-sqrt(100-4xx1xx2))/2=(-10+-sqrt92)/2=(-10+-2sqrt23)/2`

i.e. `-5-sqrt23` and `-5+sqrt23`