Precalculus Question

Question

Find a polynomial function with real coefficients that has 1 and 4i as it roots.

a. x3+x2+16x + 16

b. x3–x2–16x – 16

c. x3+x2+16x + 16

d. x3–x2–16x + 16

Answer 1

NONE of these options are correct.

Then -4i is also a solution.

THe factors are (x-1)(x+4i)(x-4i)

= (x-1)(x^2 + 16)

= x^3 - x^2 + 16x - 16

Note that x=1 is NOT a solution for polynomials A,B, and C, as the result is not zero.

For polynomial D, x=4i is NOT a solution because

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

-64i - -16 - 64i + 16 = -128i + 32

But in the polynomial above in bold, for x=4i

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

-64i - (-16) + 64i - 16 is zero

and also 1 is clearly a solution.

None of the given answers are correct, but the polynomial above in bold is correct.