Question

How do you find all the zeros of `P(X) = x^3 - 6x^2 + 4x + 16`?

Answer 1

the roots or zeros are: `x_(1,2)=1+-sqrt(5) and x=4` graph{x^3-6x^2+4x+16 [-2.603, 6.86, -1.507, 3.23]}

Explanation:

Using Zero Rational Theorem you see that the potential roots `p/q` are `{1,2,4,8,16}`.
First root that we will try will `x-4`
`(x^3 - 6x^2 + 4x + 16)/(x-4)` apply long division

`4 |bar(1 - 6 + 4 + 16`
`" " |1 + 4 - 8 - 16`
`" " |1 -2 - 4 - 0`

Thus `(x-4)` divides our polynomial without remainder and `:.` it is a factor.
`(x^3 - 6x^2 + 4x + 16)=color(brown)((x^2-2x-4))(x-4)`
Now factor the binomial `color(brown)(x^2-2x-4)`
Use quadratic formula:
`x_(1,2)=-(b+-sqrt(b^2-4ac))/(2a)=(2+- sqrt((-2)^2-4(1)(-4) ))/2`
`x_(1,2)=(2+- sqrt(4+16))/2=(2+- 2sqrt(5))/2=1+-sqrt(5)`

So the roots or zeros are: `x_(1,2)=1+-sqrt(5) and x=4`