Question

Factor the expression completely over the complex numbers.

Answer 1

Answer:

The factor form of given expression is (x-4)(x-2i)(x+2i).

Step-by-step explanation:

The given expression is

$x^3-4x^2+4x-16$

It can be written as

$f(x)=x^3-4x^2+4x-16$

According to the rational root theorem, all possible rational roots are in the form of

$\frac{a_0}{a_n}$

Where, a₀ is constant term and $a_n$ is leading coefficient.

$f(4)=4^3-4(4)^2+4(4)-16=0$

Since the value of f(x) is 0 at x=4, therefore 4 is a root of the function and (x-4) is a factor of given expression.

Use synthetic method to find the remaining factors.

$(x^3-4x^2+4x-16)=(x-4)(x^2+4)$

$(x^3-4x^2+4x-16)=(x-4)(x^2-(2i)^2)$

$(x^3-4x^2+4x-16)=(x-4)(x-2i)(x+2i)$

Therefore the factor form of given expression is (x-4)(x-2i)(x+2i).

Answer 2

Answer:

(x-4)  (x-2i) (x+2i)

Step-by-step explanation:

x^3−4x^2+4x−16

I will factor by grouping

x^3−4x^2+  4x−16

The first group is the first 2 terms.  I can factor out an x^2

x^2 (x-4) + 4x-16

The second group I can factor out a 4

x^2 (x-4) + 4(x-4)

Now I can factor out (x-4)

(x-4)  (x^2 +4)

We know that x^2 + 4 factors into +2i and -2i  because

(a^2 -b^2) = (a-b) (a+b)  but  we have (a^2 + b^2)  = (a-bi)  (a+bi)

(x-4) (x-2i) (x+2i)