SOLUTION: Hello! How do I solve this polynomial with the real number system? 2x^3+3x^2 +2x +3=0 Please help me as soon as you can! Thank you very much!

Question 1013511: Hello!
How do I solve this polynomial with the real number system?
2x^3+3x^2 +2x +3=0
Please help me as soon as you can!
Thank you very much!
Found 2 solutions by MathLover1, rothauserc:
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!
........easiest way is to factor it completely and use zero product theorem

if =>=>=> or =>-> two complex solutions
if =>=>-> one real solution
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
Using the rational zeros test on
2x^3 +3x^2 +2x +3 = 0
the constant is 3 with 1, 3 as factors
the constant of the x^3 term is 2 so we have
possible zeros are + or - 1/2 and 3/2 = -1/2, -3/2, 1/2, 3/2
let's try -3/2
2*(-3/2)^3 +3*(-3/2)^2 +2(-3/2) + 3 = 0
-27/4 +27/4 -3 +3 = 0
0 = 0
we now know that -1.5 is one of our roots
next is to do synthetic division
*************************************************************
1.5 | 2 3 2 3 |
: 3 9 16.5
:::: 2 6 11 19.5
therefore the 2nd degree polynomial is
2x^2 + 6x + 11 + (19.5 / (x - 1.5))
************************************************************
the rational root is x = -1.5
also there are two imagery roots i and -i
***********************************************************
Solve for x
2x^3 +3x^2 +2x +3 = 0
:
The left hand side factors into a product with two terms:
(2x+3) (x^2+1) = 0
:
Split into two equations:
2x+3 = 0 or x^2+1 = 0
:
Subtract 3 from both sides:
2x = -3 or x^2+1 = 0
:
Divide both sides by 2:
x = -3/2 or x^2+1 = 0
:
Subtract 1 from both sides:
x = -3/2 or x^2 = -1
:
Take the square root of both sides:
*****************************************
x = -3/2 or x = i or x = -i

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