Find a polynomial of degree #3# that has zeros, #1#, #-2#, and #3#, and in which the coefficient of #x^2# is #3#?

1 Answer
Feb 2, 2018

#f(x) = -3/2x^3+3x^2+15/2x-9#

Explanation:

Each zero #a# corresponds to a linear factor #(x-a)#.

So we can write a monic cubic polynomial with the required zeros by multiplying as follows:

#(x-1)(x+2)(x-3) = x^3-2x^2-5x+6#

Then to make the coefficient of #x^2# into #3#, multiply by #-3/2# to get:

#f(x) = -3/2x^3+3x^2+15/2x-9#