Find a fourth degree polynomial with real coefficients that has zeros of -3, 2, i that passes through (-2, 100).

Step-by-step explanation:

A fourth degree polynomial is of the form

f(x) = K(x-a)(x-b)(x-c)(x-d)

Where K is a constant and a,b,c, and d are roots of the equation.

We have three roots already: -3, 2, i

Because we are told that the coefficients are real and we have an imaginary zero, i, we need to obtain its conjugate which is = -i

So, the four roots are: a = -3, b = 2, c = i, d = -i

f(x) = K(x-(-3))(x-2)(x-i)(x-(-i))

f(x) = K(x+3)(x-2)(x-i)(x+i)

But

The polynomial passes through (-2, 100). That is, at x = -2, f(x) = 100

answer: the answer is one

step-by-step explanation:

pemdas, in this situation there is no parenthesis or exponents so you go to multiplication, 1x1=1, then division 1/1= 1 then addition 1+1=2 then subtraction 2-1=1

$119.20

step-by-step explanation:

$372.50 - 60% off is $149. $149-20% off is $119.20