n=​3; 4 and 4 i are​ zeros; f(1) = -51

Question

nth degree polynomial function

Answer 1

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

f(x) = a(x - 4)(x² + 16)

f(x) = a(x³ - 4x² + 16x - 48)

-51 = a(1 - 4 + 16 - 48)

Can you solve for a and answer?

Answer 2

Hello Nathan,

Complex roots always occur in CONJUGATE PAIRS. So if 4i is a root then so is -4i. The third root is given as 4.

Write the roots in factor form: ( x - 4) (x + 4i) (x - 4i).= f(x)

Multiply the factors together. It will be easier if you multiply the conjugates first.

So FOIL ( x + 4i ) ( x - 4i ) = x ² -4i x + 4i x -16 i ² since i ² = -1 then

SIMPLIFIED we get ( x ² + 16 )

Multiply using FOIL again ( x ² +16 ) ( x - 4) = x ³ - 4 x ² + 16 x - 64 =f(x)

Check our function value substitute 1 for every x in the function f(1) =(1)³ - 4 (1)² +16(1) -64 = -51 Yippeee