Question

Write a polynomial f (x) that satisfies the given conditions. Polynomial of lowest degree with zeros of -4 (multiplicity 3), 1 (multiplicity 1), and with f(0) = 320.​

Answer 1

Answer:

Step-by-step explanation:

Polynomial f(x) has the following conditions: zeros of -4 (multiplicity 3), 1 (multiplicity 1), and with f(0) = 320.

The first part zeros of -4 means (x+4) and multiplicity 3 means (x+4)^3.

The second part zeros of 1 means (x-1) and multiplicity 1 means (x-1).

The third part f(0) = 320 means substituting x=0 into (x+4)^3*(x-1)*k =320

(0+4)^3*(0-1)*k = 320

-64k = 320

k = -5

Combining all three conditions, f(x)

= -5(x+4)^3*(x-1)

= -5(x^3 + 3*4*x^2 + 3*4*4*x + 4^3)(x-1)

= -5(x^4 + 12x^3 + 48x^2 + 64x - x^3 - 12x^2 - 48x - 64)

= -5(x^4 + 11x^3 + 36x^2 + 16x -64)

= -5x^3 -55x^3 - 180x^2 - 80x + 320

Answer 2

Answer:

Step-by-step explanation:

-4 is a root for 3 times and 1 is root for once

so (x+4)^3 * (x-1) is part of f(x)

the constant term there is 4^3*(-1)=-64

so there is a multiplier of 320/-64=-5

f(x) = -5 * (x+4)^3 * (x-1)