# college algebra

posted by on .

I'm stuck on 2 problems I know the answers but need to learn to show the work.

3x^3+4x^2+6=x. (-2,1/3, i√8/3)

• college algebra - ,

X^4+x^3=16-8x-6x^2

x^4 + x^3 + 6x^2 + 8x - 16 = 0

you must know the factor theorem to do these

look at the constant of 16
if there are factors of the type ( ?x + c) , c must be a factor of 16
so try ±1, ±2, ±4, ... that is, try factors of 16
e.g.
let x = 1 , we get
1 + 1 + 6 + 8 - 16 = 0 ahhhh, so x-1 is a factor, that was lucky
try some more:
x = 2, we get
16 + 8 + 24 + 16 - 16 ≠ 0
x = -2
16 - 8 + 24 - 16 - 16 = 0 , aahhh again, so x+2 is a factor

now do a long algebraic division by first x-1, and then x+2
both must divide evenly.
I got
x^4 + x^3 + 6x^2 + 8x - 16 = (x-1)(x+2)(x^2 + 8)

so x = 1 , -2 or
x^2 = -8 = 2√-2 = ±2√2 i , yours should have been ±√8 i

2nd:
3x^3 + 4x^2 - x + 6 = 0
try x = ±1, ±2, ±3

x=1 ---> 1 + 4 - 1 + 6 ≠0
x=-1 --> -3 + 4 + 1 + 6 ≠ 0
x = 2 --> 24 + 16 ...... ≠ 0
x = -2 --> -24 + 16 + 2 + 6 = 0 ..... finally, x+2 is a factor
division:
3x^3 + 4x^2 - x + 6 = (x+2)(3x^2 - 2x + 3)

so x = -2
or
x = (2 ±√-32)/6
= ( 2 ± 4√2 i)/6
= (1 ± 2√2 i)/3