# Problem 6_5 (C- part 1 & E- part 2)

Author:
angela

## 6_5

Solve the cubic equations: (1) . (2) .

## Solution:

1- It is very clear that x=1 is a solution because 1-6+11-6=0. If we divide the cubic polynomial by (x-1), we get . We can then factor to get (x-2)(x-3), which means 2 and 3 are also solutions. 2- I first subtracted from both sides. This gave me . Then, I took the cubed root of both sides. This left me with (x-2)=-(x-1). I then distributed a -1 on the right side, which left me with x-2 = -x+1. I added 2 to both sides and then added x to both sides. I was left with 2x=3. I then divided by 2 in order to get a root of 3/2. Define: then, with solutions Since --> When y=1, x=3/2. When ,= Thus, . When ,