Write a polynomial f (x) that meets the given conditions. Degree 3 polynomial with integer coefficients with zeros -7i and

Answer and Step-by-step explanation:

Since it is a degree 3 polynomial, then it should have 3 solutions. But only one is given -7i which is an imaginary number. Since -7i is part of the solutions, then definitely, +7i is also part of the solutions. Since we are not given the last root, I would choose a suitable number in order to explain the process. Let us choose 1. Therefore, our roots are now:

±7i and 1.

In order to find the polynomial, we have:

x = +7i or –7i or 1, this means:

f(x) = (x – 7i)(x + 7i)(x – 1)

f(x) = [x² +7ix – 7ix – (7i)²](x – 1) = [x² – 49(–1)](x – 1) = (x² + 49)(x – 1) = x³ – x² + 49x – 49

d

step-by-step explanation:

you didn't post any answer choices, so its not 100% clear what your teacher is looking for

however, log(15/7) is equivalent to log(15) - log(7)

i'm using the rule log(x/y) = log(x) - log(y)