Factor f(x) into linear factors given that k is a zero of f(x).
f(x)=x^3+(12-4i)x^2+(32-48i)x-128i, k=4i
In completely factored form, f(x) =____
(factor completely. simplify your answer)
Since you are telling me that 4i = k
then
f(x)=x^3+(12-4i)x^2+(32-48i)x-128i
=x^3 + (12 - k)x^2 + (32 - 12k) - 32k
and since k is a zero, then (x - k) is a factor
by synthetic division:
f(x)=x^3+(12-4i)x^2+(32-48i)x-128i
= (x-k)(x^2 + 12x + 32)
= (x - 4i)(x^2 + 12x + 32)
To factor the polynomial f(x) into linear factors, we will make use of the fact that if k is a zero of f(x), then (x - k) is a factor of f(x).
Given that k = 4i is a zero of f(x), we can rewrite f(x) as:
f(x) = (x - 4i)(ax^2 + bx + c)
where a, b, and c are constants to be determined.
To find a, b, and c, we can perform polynomial division or long division. However, since synthetic division is most efficient in this case, we will use it.
Performing synthetic division with k = 4i, we get:
4i | 1 12-4i 32-48i -128i
------------------------------------------
1 -4i 16 0
The quotient obtained from synthetic division is: x^2 - 4ix + 16.
Therefore, f(x) = (x - 4i)(x^2 - 4ix + 16).
To factor the quadratic term x^2 - 4ix + 16 further, we can apply the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For the quadratic term, a = 1, b = -4i, and c = 16.
Plug in these values and simplify:
x = (4i ± √((4i)^2 - 4(1)(16))) / (2(1))
= (4i ± √(-16 - 64)) / 2
= (4i ± √(-80)) / 2
= (4i ± 4i√5) / 2
= 2i ± 2i√5
Therefore, the completely factored form of f(x) is:
f(x) = (x - 4i)(x - (2i + 2i√5))(x - (2i - 2i√5))
Simplifying further, we can combine the complex conjugates:
f(x) = (x - 4i)(x - 2i - 2i√5)(x - 2i + 2i√5)
This is the completely factored form of f(x).