college algebra
posted by Anonymous on .
I'm stuck on 2 problems I know the answers but need to learn to show the work.
X^4+x^3=168x6x^2. Answer: 2,1,+√8
3x^3+4x^2+6=x. (2,1/3, i√8/3)
Please help so I know how to figure these equations out.A big THANKS

X^4+x^3=168x6x^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 x1 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 x1, and then x+2
both must divide evenly.
I got
x^4 + x^3 + 6x^2 + 8x  16 = (x1)(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
I don't see how you got your answers.