Question

Write a polynomial f(x) that satisfies the given conditions. Polynomial of lowest degree with zeros of -2 (multiplicity 3), 3 (multiplicity 1), and with f(0) = 120.​

Answer 1

Answer:

Step-by-step explanation:

Polynomial f(x) has the following conditions: zeros of -2 (multiplicity 3), 3 (multiplicity 1), and with f(0) = 120.

The first part zeros of -2 means (x+2) and multiplicity 3 means (x+2)^3.

The second part zeros of 3 means (x-3) and multiplicity 1 means (x-3).

The third part f(0) = 120 means substituting x=0 into (x+2)^3*(x-3)*k =120

(0+2)^3*(0-3)*k = 120

-24k = 120

k = -5

Combining all three conditions, f(x)

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

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

= -5(x^4 + 6x^3 + 12x^2 + 8x - 3x^3 - 18x^2 - 36x - 24)

= -5(x^4 + 3x^3 - 6x^2 - 28x -24)

= -5x^4 - 15x^3 + 30x^2 + 140x + 120

Answer 2

Answer:

Step-by-step explanation:

-2 is a root for 3 times and 3 is root for once

so (x+2)^3 * (x-3) is part of f(x)

the constant term there is 2^3*(-3)=-24

so there is a multiplier of 120/-24=-5

f(x) = 5 * (x+2)^3 * (x-3)