Find the polynomial with integer coefficients that satisfies the following conditions

Question

Degree of polynomial: 3

Zeros: 4, 4i

Constant coefficient: -192

 

A.3x^212x2+48x-192

B.x^3-6x^2-48x-192

C.4x^3-3x^2+48x-192

D.x^3-12x^2+48x-192

E.x^3+12x^2+48+192

F. None of the above

Answer 1

Complex numbers have conjugates.  So the other conjugate will give us an additional zero.

 

x₁ = 4

x₂ = 4i

x₃ = -4i

 

 

Using these roots, we can transformed into factors for the polynomial f(x).

 

f(x) = C(x - 4)(x - 4i)(x + 4i)

 

where C is the arbitrary constant.

 

 

Expanding, we get

 

f(x) = C(x - 4)(x² + 16)

 

f(x) = C(x³ - 4x² + 16x - 64)

 

 

Now we use the constant coefficient to find C.  The constant coefficient is the y-intercept.  That is  f(0)=-192.

 

-192 = C(-64)

 

3 = C

 

 

Plugging in this value of C,

 

f(x) = 3(x³ - 4x² + 16x - 64)

 

f(x) = 3x³ - 12x² + 48x - 192

 

 

This matches none of the answer choices.  However, answer choice A is not written correctly, so you will need to fix that and compare the choice to the answer I calculated for you.

Answer 2

x = 4   ==>   x - 4 = 0

x = 4i   ==>   x^2 = 16i^2 = 16(-1) = -16   ==>   x^2 + 16 = 0

 

(x - 4) * (x^2 + 16) = 0

x^3 + 16x - 4x^2 - 64 = 0

x^3 - 4x^2 + 16x - 64 = 0

 

To get the constant to be -192, multiply the whole equation by -192 / -64 = 3

 

3 * (x^3 - 4x^2 + 16x - 64) = 0

3x^3 - 12x^2 + 48x - 192 = 0