Find a polynomial (with real coefficients) of degree 4 with zeros at -3 (multiplicity of 2) and -2i, and with 6 as the coefficient of x

Step-by-step explanation:

If -2i is a root, then so is +2i, so

y = a(x+3)^2(x^2+4)

= a(x^4+6x^3+13x^2+24x+36)

Since we want 24a = 6, our final polynomial is

y = 1/4 (x^4+6x^3+13x^2+24x+36)

No, the points are in a linear pattern.

Step-by-step explanation:

Residual = actual value - predicted value

So our residual points are:

1, -2.4

2, -4.9

3, -6.2

4, -6

5, -7.9

As evidenced in the image, the points follow a linear pattern.