Find a polynomial of degree 3 with real coefficients and zeros of -3,-1, and 4, for which f(-2)=-24

We want to find a polynomial

f(x) = a x³ + b x² + c x + d

such that the roots of f are x = -3, x = -1, and x = 4, and f(x) takes on a value of -24 when x = -2.

The factor theorem for polynomials tells us that we can factorize f(x) as

a x³ + b x² + c x + d = a (x + 3) (x + 1) (x - 4)

Expand the right side:

(x + 3) (x + 1) (x - 4) = x³ - 13x - 12

So we have

a x³ + b x² + c x + d = a x³ - 13a x - 12a

In order for both sides to be equal, the coefficients of both polynomials on terms of the same degree must be equal. This means

a = a (of course)

b = 0 (there is no x² term on the right)

c = -13a

d = -12a

We also have that f (-2) = -24, which means

f (-2) = a (-2 + 3) (-2 + 1) (-2 - 4)

-24 = 6a

a = -4

which in turn tells us that c = 52 and d = 48.

So we found

f(x) = -4x³ + 52x + 48

answer: 12   they all have at least 28 days

step-by-step explanation: haha