exact roots of 3x^3+9x^2+10x+4=0??

Question

thank you!

Answer 1

f (-1) = 0 thus x - (-1) = x + 1 is a factor.
To find other factors, use synthetic division:-
__|3__9__10__4
- 1|__- 3__- 6_- 4
__|3__6___4__0

(x + 1)(3x² + 6x + 4) = 0

x = [ (- 6) ± √(36 - 48) ] / 6 , x = -1
x = [(- 6) ± √ (- 12) ] / 6 , x = - 1
x = [ (- 6) ± i 2 √3) ] / 6 , x = - 1

Answer 2

If there are any rational roots, we can (absolutely, positively) say that they will be of the form "a factor of 4"/"a factor of 3"...
Trial and error leads us to find the -1 is a root.
(x+1)(3x^2+6x+4)=0
Now, using the quadratic formula, we can find the other 2 roots.
x=i*sqrt(1/3)-1, -i*sqrt(1/3)-1, -1

Answer 3

3x^3+9x^3+10x+4=0
3x^3+3x^2+6x^2+6x+4x+4=0
3x^2(x+1)+6x(x+1)+4(x+1)=0
(x+1)(3x^2+6x+4)=0
because 3x^2+6x+4=3x^2+6x+3+1=3(x+1)^2+1>0
so x+1=0
x= -1!!!!