Find real and complex zeros of the polynomial

[hasty osprey]

P(x) = x^4 − 4x^3 + 8x^2 − 16x + 16
x=

[modest dagger]

you could use factor theorem to show that x=2 is a zero of the polynomial

then factor out (x-2) giving (x-2)(x^3-2x^2+4x-8)

use factoring by grouping to factorise the cubic

and you'll find you have a zero of multiplicity 3 at x=2 and a zero of multiplicity 1 at x=-4

[noble helm]

Isn't factor should be 4?

[weak wraith]

No

x=2 is a root

[noble helm]

But 4 can divide all numbers evenly

[weak wraith]

The theorem says

That the root must be a divisor of 16

In fact x = 2 is a double root

Because P(2) = 0
And P'(2) = 0

Where P' is the derivative of P

So
$P(x) = (x-2)^2 Q(x)$

[verbal citrus]

Daddy_314

[weak wraith]

With Q of degree 2

[modest dagger]

4 divides the coefficients that you see evenly

but that doesnt mean it is a factor

you're also missing that the coefficient of x^4 is 1

and 4 doesnt divide that evenly either

[fluid badger]

2 is only a zero of multiplicity 2 then P(x) = (x-2)² Q(x), but Q is a quadratic who hasn't any real root.

[fluid badger]

4 can divide all numbers except x^4