Question

What is the third–degree polynomial function such that f(0) = –18 and whose zeros are 1, 2, and 3?

f(x) = 2x3 – 12x2 + 18x – 18

f(x) = 3x3 + 18x2 – 33x – 18

f(x) = 2x3 + 12x2 – 18x – 18

f(x) = 3x3 – 18x2 + 33x – 18

Answer 1

f(x) = 3x3 – 18x2 + 33x – 18

Answer 2

How?

Answer 3

are u sure that's the answer?

Answer 4

Oh i got it.. :D

Answer 5

thanks

Answer 6

By elimination:
Using Descartes' Rule of Signs only 1st and 4th equation makes sense
Plugging in the number 1, only 4th equation makes sense, eliminating 1st

Or more directly:
f(x)=3x3 – 18x2 + 33x – 18
=(x-1)(3x^2 - 15x + 18)
= (x-1)3(x^2 - 5x + 6)
=3(x-1)(x-2)(x-3)

f(x) = 0 when x = 1,2 or 3

Answer 7

most factoring techniques I know weren't working so I started with (x-1) as common factor and everything worked out

Answer 8

I could have also used The Rational Root Theorem, which would have given me these rational roots. DRoS shows that roots are positive and real, another hint. Only factors of 6 are 1,2,3 and 6 as possible roots (based on RRT) of these 3 turned out to be the roots required.