Question

Find a cubic function with the given zeros.

7, -3, 2

Answer 1

There an an infinite number of cubic functions you can generate
In general

\[y = k(x - a)(x-b)(x-c)\] if a cubic with zeroes at x = a, b and c
k is just a real, non zero constant (any number just not zero)

Answer 2

here are the answer choices
A. x^3 - 6x^2 -13x - 42
B. x^3 - 6x^2 +13x +42
C. x^3 -6x^2 -13x + 42
D. x^3 + 6x^2 -13x +42

Answer 3

well do as I said and expand the parentheses. use k = 1 to get a monic cubic.

Answer 4

actually there is an easier way

Answer 5

you can use sum, product of roots

Answer 6

i dont know how to expand them

Answer 7

how do you do that

Answer 8

For monic cubic with roots 7,-3, and 2, you can write this as:
y = (x - 7)(x+3)(x-2)

Expand
(x-7)(x+3)(x-2) = (x^2 +3x - 7x -21)(x-2) = (x^2 -4x -21)(x-2)
= x^3 - 2x^2 - 4x^2 + 8x -21x +42
= X^3 - 6x^2 -13x + 42

Answer 9

thank you!

Answer 10

so its c