Question

Question #ba224

Answer 1

two real roots `x=3and2`
two imaginary roots `x=+-i`

Explanation:

before searching for the zeros in f(x)
we need to know there are two conditions
1. four real roots
2.two real roots & two imaginary roots
imaginary roots always be pair!

graph{y=x^4-5x^3+7x^2-5x+6 [-8.89, 8.89, -4.44, 4.45]}
let `f(x)=x^4-5x^3+7x^2-5x+6`
if x=3 `f(3)=81-135+63-15+6=0`
so `f(x)`can be divided by `(x-3)`

`(x^4-5x^3+7x^2-5x+6)/(x-3)=x^3-2x^2+x-2`

the next step also use by the same method

graph{y=x^3-2x^2+x-2 [-8.89, 8.89, -4.44, 4.45]}
let`g(x)=x^3-2x^2+x-2`
if `x=2`
`g(x)=8-8+2-2=0`

`g(x)/(x-2)=x^2+1`
exist two imaginary roots `+-i` in the equation `x^2+1`