Write a polynomial function of least degree with integral coefficients that has the given zeros. -5, 3i

Solving for the polynomial function of least degree with integral coefficients whose zeros are -5, 3i

We have:
x = -5

Then x + 5 = 0

Therefore one of the factors of the polynomial function is (x + 5)

Also, we have:
x = 3i
Which can be rewritten as:
x = Sqrt(-9)
Square both sides of the equation:
x^2 = -9
x^2 + 9 = 0

Therefore one of the factors of the polynomial function is (x^2 + 9)

The polynomial function has factors: (x + 5)(x^2 + 9)
= x(x^2 + 9) + 5(x^2 + 9)

= x^3 + 9x + 5x^2 = 45

Therefore, x^3 + 5x^2 + 9x – 45 = 0

f(x) = x^3 + 5x^2 + 9x – 45

The polynomial function of least degree with integral coefficients that has the given zeros, -5, 3i is f(x) = x^3 + 5x^2 + 9x – 45

a

step-by-step explanation:

answer: 1,380

step-by-step explanation:

1: you turn %15 into a decimal by moving the decimal point two places to the left.

2: and then you multiply .15 by 1,200 , which will get you 180.00

3: from there you will add 180 to 1,200

4: which will get you 1,380

hope it *im kind of slow so idk if my answer is right*