Find the polynomial f(x) of degree three that has zeroes at 1, 2, and 4 such that f(0) = -16.?

Question

1.Find the polynomial f(x) of degree three that has zeroes at 1, 2, and 4 such that f(0) = -16.

2.For the following equation, find the interval(s) where f(x) < 0.
f(x) = 1 Divided by x^2-2x-8

3.The polynomial f(x) divided x - 3 results in a quotient of x^2+3x-5 with a remainder of 2. Find f(3).

4.Express the following statement as a formula with the value of the constant of proportionality determined with the given conditions: w varies directly as x and inversely as the square of y. If x = 15 and y = 5, then w = 36.

5.Find the quotient and remainder of f(x) = x^4 - 2 divided by p(x) = x - 1.

Answer 1

Too many questions. Here's the answer for the one in the title.

Zeroes at 1, 2, 4 means (x-1), (x-2), and (x-4) are factors. So f(x) = c(x-1)(x-2)(x-4) for some constant c. Multiplying the terms out gives c(x^3 -7x^2 + 14x - 8). Then f(0) = -8c, so c = 2 gives f(0) = -16. That means f(x) = 2x^3 - 14x^2 + 28x - 16.

Answer 2

the answer is probable option a(i.e.-a million) i'm no longer positive a thanks to do it, yet that is how i have been given to the answer: f(-a million) = -a million = -a million(-a million)^2 f(2) = -4 = -a million(2)^2 f(-3) = -9 = -a million(-3)^2 f(4) = -16 = -a million(-4)^2 So f(a million) ought to equivalent to -a million(a million)^2 = -a million wish i'm perfect