Factor the polynomial function over the complex numbers. f(x)=x^4−x^3−2x−4 enter your answer in the box.

x^4−x^3−2x−4 = (x - √2 i)*(x + √2 i)*(x-2)*(x+1)

Step-by-step explanation:

Given

f(x)=x^4−x^3−2x−4  

To factorize it, first reorder f(x) as follows:

x^4 −4  − x^3− 2x       (x^4 −4 is a difference of two squares)

(x^2 - 2)*(x^2 + 2) - x*(x^2 + 2)        (take x^2 + 2 as common factor)

(x^2 + 2)*(x^2 - 2 - x)      (reordering and equating to zero)

(x^2 + 2)*(x^2 - x - 2) = 0

We have two options

(x^2 + 2) = 0 <=> x^2 = -2 <=> x = √2 i or x = -√2 i

Or

(x^2 - x - 2) = 0     (using quadratic formula)

Finally f(x)=x^4−x^3−2x−4 = (x - √2 i)*(x + √2 i)*(x-2)*(x+1)

idk but i found this "the graph of y = f(x-2) is the graph of y = f(x) translated two units rightward.  

so, since the point (9,2) lies on the graph of y = f(x), the point (11,2) lies on the graph of y = f(x-2)."

step-by-step explanation:

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